Equiconvex lens Plano convex lens Equi concave lens Plano concave lens
R1 = R and R2 = −R R1 = ∞, R2 = −R R1 = −R , R2 = +R R1 = ∞ , R2 = R
f = R f = R f = − R f = R
2(µ − 1) (µ − 1) 2(µ − 1) 2(µ − 1)
for µ = 1.5 , f = R for µ = 1.5 , f = 2R for µ = 1.5 f = −R for µ = 1.5, f = −2R
(5) Lens in a liquid
Focal length of a lens in a liquid (fl ) can be determined by the following formula
fl = (a μg − 1) (Lens is supposed to be made of glass).
fa (l μg − 1)
Note : ≅Focal length of a glass lens (µ = 1.5) is f in air then inside the water it’s focal length is 4f.
≅ In liquids focal length of lens increases (↑) and it’s power decreases (↓).
(6) Opposite behaviour of a lens
In general refractive index of lens (µ L ) > refractive index of medium surrounding it (µ M ).
µL > µM µL < µM µL = µM
(7) Lens formula and magnification of lens
(i) Lens formula : 1 11 ; (use sign convention)
f =v−u
(ii) Magnification : The ratio of the size of the image to the size of object is called magnification.
(a) Transverse magnification : m= I = v = f = f −v (use sign convention while solving the problem)
O u +u f
f
I v2 − v1 dv v 2 = f 2 f − v 2
O u2 − u1 du u + f
(b) Longitudinal magnification : m= = . For very small object m= = f u =
(c) Areal magnification : ms = Ai = m2 = f f u 2 , (Ai = Area of image, Ao = Area of object)
Ao +
(8) Relation between object and image speed
If an object move with constant speed (Vo ) towards a convex lens from infinity to focus, the image will move
f 2
+
slower in the beginning and then faster. Also Vi = f u . Vo
(9) Focal length of convex lens by displacement method
(i) For two different positions of lens two images (I1 and I 2 ) of an object is formed at the same location.
(ii) Focal length of the lens f = D2 − x2 x x
4D = m1 − m2 D > 4f
Object
where m1 = I1 and m2 = I2 O I2
O O I1
Screen
(iii) Size of object O = I1. I 2
(10) Cutting of lens
(i) A symmetric lens is cut along optical axis in two equal parts. Intensity of image formed by each part will be
same as that of complete lens.
(ii) A symmetric lens is cut along principle axis in two equal parts. Intensity of image formed by each part will
be less compared as that of complete lens.(aperture of each part is 1 times that of complete lens)
2
⇒ f, P
2f 2f ⇒
f, P P/2 P/2
f, P
(11) Combination of lens
(i) For a system of lenses, the net power, net focal length and magnification given as follows :
P = P1 + P2 + P3 .......... , 1 = 1 + 1 + 1 + ..........., m = m1 × m2 × m3 ×............
F f1 f2 f3
(ii) In case when two thin lens are in contact : Combination will behave as a lens, which have more power or
lesser focal length.
11 1 F = f1 f2 and P = P1 + P2
F = f1 + f2 ⇒ f1 + f2
(iii) If two lens of equal focal length but of opposite nature are in contact then combination will behave as a
plane glass plate and Fcombination = ∞
(iv) When two lenses are placed co-axially at a distance d from each other then equivalent focal length (F).
111 d and P = P1 + P2 − dP1P2 f1 f2
F = f1 + f2 − f1 f2 d
(v) Combination of parts of a lens :
⇒ and and
F = f/2 F=∞
F =f F=f f
f
(12) Silvering of lens
On silvering the surface of the lens it behaves as a mirror. The focal length of the silvered lens is 1 = 2 + 1
F fl fm
where fl = focal length of lens from which refraction takes place (twice)
fm = focal length of mirror from which reflection takes place.
(i) Plano convex is silvered
⇒+ ⇒+
F fl fm F fl fm
fm = R , fl = R so F = R fm = ∞, fl = R so F = R
2 (µ − 1) 2µ (µ − 1) 2(µ − 1)
(ii) Double convex lens is silvered
Since fl = 2 R 1) , fm = R ⇒ +
(µ − 2
F fl fm
So F R
= 2(2µ − 1)
Note : ≅ Similar results can be obtained for concave lenses.
(13) Defects in lens
(i) Chromatic aberration : Image of a white object is coloured and blurred because µ (hence f) of lens is
different for different colours. This defect is called chromatic aberration.
White Real Violet µV > µ R so fR > fV
light FV FR Mathematically chromatic aberration = f R − fV = ωfy
fV ω = Dispersion power of lens.
fR fy = Focal length for mean colour = fR fV
Removal : To remove this defect i.e. for Achromatism we use two or more lenses in contact in place of single lens.
Mathematically condition of Achromatism is : ω1 + ω2 =0 or ω1 f2 = −ω 2 f1
f1 f2
Note : ≅ Component lenses of an achromatic doublet cemented by canada blasam because it is
transparent and has a refractive index almost equal to the refractive of the glass.
(ii) Spherical aberration : Inability of a lens to form the point image of a point object on the axis is called
Spherical aberration.
In this defect all the rays passing through a lens are not focussed at a single point and the image of a point
object on the axis is blurred.
Marginal rays
Paraxial ray F′ F F′ F
Removal : A simple method to reduce spherical aberration is to use a stop before and infront of the lens. (but
this method reduces the intensity of the image as most of the light is cut off). Also by using plano-convex lens, using
two lenses separated by distance d = F – F ', using crossed lens.
Note : ≅ Marginal rays : The rays farthest from the principal axis.
Paraxial rays : The rays close to the principal axis.
≅ Spherical aberration can be reduced by either stopping paraxial rays or marginal rays, which
can be done by using a circular annular mask over the lens.
≅ Parabolic mirrors are free from spherical aberration.
(iii) Coma : When the point object is placed away from the principle axis and the image is received on a
screen perpendicular to the axis, the shape of the image is like a comet. This defect is called Coma.
It refers to spreading of a point object in a plane ⊥ to principle axis.
Image of P P′
Axis
P
Removal : It can be reduced by properly designing radii of curvature of the lens surfaces. It can also be
reduced by appropriate stops placed at appropriate distances from the lens.
(iv) Curvature : For a point object placed off the axis, the image is spread both along and perpendicular to
the principal axis. The best image is, in general, obtained not on a plane but on a curved surface. This defect is
known as Curvature.
Removal : Astigmatism or the curvature may be reduced by using proper stops placed at proper locations
along the axis.
(v) Distortion : When extended objects are imaged, different portions of the object are in general at different
distances from the axis. The magnification is not the same for all portions of the extended object. As a result a line
object is not imaged into a line but into a curve.
Object Distorted images
(vi) Astigmatism : The spreading of image (of a point object placed away from the principal axis) along the
principal axis is called Astigmatism.
Concepts
If a sphere of radius R made of material of refractive index µ 2 is placed in a medium of refractive index µ1 , Then if the object is
placed at a distance µ µ1 µ1 R from the pole, the real image formed is equidistant from the sphere.
2−
µ1 µ2
O P1 P2 I
µ2
x 2x x
The lens doublets used in telescope are achromatic for blue and red colours, while these used in camera are achromatic for violet
and green colours. The reason for this is that our eye is most sensitive between blue and red colours, while the photographic
plates are most sensitive between violet and green colours.
Position of optical centre
Equiconvex and equiconcave Exactly at centre of lens
Convexo-concave and concavo-convex Outside the glass position
Plano convex and plano concave On the pole of curved surface µ1
Composite lens : If a lens is made of several materials then µ2
Number of images formed = Number of materials used µ3
Here no. of images = 5
µ4
µ5
Example
Example: 18 A thin lens focal length f1 and its aperture has diameter d. It forms an image of intensity I. Now the central
Example part of the aperture upto diameter d/2 is blocked by an opaque paper. The focal length and image intensity
will change to
Solution: (d) [CPMT 1989; MP PET 1997; KCET 1998]
Example: 19 (a) f and I (b) f and I (c) 3f and I (d) f and 3I
Solution: (a) 2 2 4 4 2 4
Example: 20 Centre part of the aperture up to diameter d is blocked i.e. 1 th area is blocked A = πd 2 . Hence
2 4 4
Solution: (a)
Example: 21 remaining area A′ = 3 A. Also, we know that intensity ∝ Area ⇒ I′ = A′ = 3 ⇒ I′ = 3 I .
4 I A 4 4
Focal length doesn't depend upon aperture.
The power of a thin convex lens (a µg = 1.5) is + 5.0 D. When it is placed in a liquid of refractive index a µl ,
then it behaves as a concave lens of local length 100 cm. The refractive index of the liquid a µl will be
(a) 5 / 3 (b) 4 / 3 (c) 3 (d) 5 / 4
By using
fl = a µg −1 ; where l µg = µg 1.5 and fa = 1 = 1 m = 20 cm
fa l µg −1 µl = P 5
µl
⇒ −100 = 1.5 −1 ⇒ µl = 5 / 3
20 1.5 −1
µl
A double convex lens made of a material of refractive index 1.5 and having a focal length of 10 cm is
immersed in liquid of refractive index 3.0. The lens will behave as
[NCERT 1973]
(a) Diverging lens of focal length 10 cm (b) Diverging lens of focal length 10 / 3 cm
(c) Converging lens of focal length 10 / 3 cm (d) Converging lens of focal length 30 cm
By using fl = aµg −1 ⇒ fl = 1.5 − 1 ⇒ fl = −10 cm (i.e. diverging lens)
fa l µg −1 + 10
1.5 − 1
3
Figure given below shows a beam of light converging at point P. When a concave lens of focal length 16 cm is
introduced in the path of the beam at a place O shown by dotted line such that OP becomes the axis of the
lens, the beam converges at a distance x from the lens. The value x will be equal to [AMU (Med.) 2002]
(a) 12 cm
(b) 24 cm P
O 12 cm
(c) 36 cm
P' P
(d) 48 cm x
Solution: (d) From the figure shown it is clear that 12cm
For lens : u = 12 cm and v = x = ?
By using 1 = 1 − 1
f v u
⇒ 1 = 1 − 1 ⇒ x = 48 cm.
+ 16 x + 12
Example: 22 A convex lens of focal length 40 cm is an contact with a concave lens of focal length 25 cm. The power of
Solution: (a)
Example: 23 combination is [IIT-JEE 1982; AFMC 1997; CBSE PMT 2000]
Solution: (b)
(a) – 1.5 D (b) – 6.5 D (c) + 6.5 D (d) + 6.67 D
Example: 24
Solution: (b) By using 111 ⇒ 11 1
Example: 25 F = f1 + f2 F = + 40 + − 25
Solution: (a)
⇒ F = − 200 cm , hence P = 100 = 100 = −1.5 D
3 f (cm) − 200 / 3
A combination of two thin lenses with focal lengths f1 and f2 respectively forms an image of distant object at
distance 60 cm when lenses are in contact. The position of this image shifts by 30 cm towards the combination
when two lenses are separated by 10 cm. The corresponding values of f1 and f2 are [AIIMS 1995]
(a) 30 cm, – 60 cm (b) 20 cm, – 30 cm (c) 15 cm, – 20 cm (d) 12 cm, – 15 cm
Initially F = 60 cm (Focal length of combination)
Hence by using 111 11 1 ⇒ f1 f2 ......(i)
F = f1 + f2 ⇒ f1 + f2 = 60 f1 + f2 ......(ii)
Finally by using 1 = 1 + 1 − d where F ′ = 30 cm and d = 10 cm ⇒ 1 = 1 + 1 − 10 …..(iii)
F′ f1 f2 f1 f2 30 f1 f2 f1 f2
From equations (i) and (ii) f1 f2 = − 600.
From equation (i) f1 + f2 = −10
Also, difference of focal lengths can written as f1 − f2 = (f1 + f2 )2 − 4 f1 f2 ⇒ f1 − f2 = 50 …..(iv)
From (iii) × (iv) f1 = 20 and f2 = −30
A thin double convex lens has radii of curvature each of magnitude 40 cm and is made of glass with refractive
index 1.65. Its focal length is nearly [MP PMT 1997]
(a) 20 cm (b) 31 cm (c) 35 cm (d) 50 cm
By using f = R ⇒ f = 40 = 30.7 cm ≈ 31cm.
2(µ − 1) 2(1.65 − 1)
A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5).
The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the
glass. The line PQ cuts the surface at a point O and PO = OQ . The distance PO is equal to
(a) 5 R (b) 3 R (c) 2 R [MP PMT 1994; Haryana CEE 1996]
(d) 1.5 R
By using µ2 − µ1 = µ2 − µ1
v u R
Where µ1 = 1, µ 2 = 1.5, u = – OP, v = OQ Q
1.5 1 1.5 − 1 1.5 1 0.5 PO
OQ − OP OP + OP = R
Hence − = (+ R) ⇒
⇒ OP = 5 R
Example: 26 The distance between an object and the screen is 100 cm. A lens produces an image on the screen when
Solution: (b) placed at either of the positions 40 cm apart. The power of the lens is
Example: 27 [SCRA 1994]
Solution: (d) (a) 3 D (b) 5 D (c) 7 D (d) 9 D
Example: 28 By using f = D2 − x2 ⇒ f = 100 2 − 40 2 = 21cm
Solution: (a) 4D 4 × 100
Example: 29
Solution: (a) Hence power P = 100 = 100 ≈ + 5D
Example: 30 21
F (cm)
Shown in figure here is a convergent lens placed inside a cell filled with a liquid. The lens has focal length +20
cm when in air and its material has refractive index 1.50. If the liquid has refractive index 1.60, the focal
length of the system is Liquid [NSEP 1994; DPMT 2000]
(a) + 80 cm Lens
(b) – 80 cm
(c) – 24 cm
(d) – 100 cm
Here 1 = (1.6 − 1) 1 − 1 = −3 .......(i)
f1 20 100
∞
1 = (1.5 − 1) 1 − 1 = 1 .......(ii) ⇒+ +
f2 20 − 20 20 f3
1 = (1.6 − 1) − 1 − 1 = −3 .......(iii) F f1 f2 [SCRA 1998]
f3 20 ∞ 100
By using 1 111 ⇒ 1 = −3 + 1 − 3 ⇒ F = −100 cm
F= f1 + f2 + f3 F 100 20 100
A concave lens of focal length 20 cm placed in contact with a plane mirror acts as a
(a) Convex mirror of focal length 10 cm (b) Concave mirror of focal length 40 cm
(c) Concave mirror of focal length 60 cm (d) Concave mirror of focal length 10 cm
By using 121
F = fl + fm
fl 20 ⇒+
2 2
Since fm = ∞ ⇒ F = = = 10 cm
(After silvering concave lens behave as convex mirror) F Fe Fm
A candle placed 25 cm from a lens, forms an image on a screen placed 75 cm on the other end of the lens.
The focal length and type of the lens should be [KCET (Med.) 2000]
(a) + 18.75 cm and convex lens (b) – 18.75 cm and concave lens
(c) + 20.25 cm and convex lens (d) – 20.25 cm and concave lens
In concave lens, image is always formed on the same side of the object. Hence the given lens is a convex lens
for which u = – 25 cm, v = 75 cm.
By using 1 = 1 − 1 ⇒ 1 = 1 − 1 ⇒ f = + 18.75 cm.
f v u f
(+ 75) (− 25)
A convex lens forms a real image of an object for its two different positions on a screen. If height of the image
in both the cases be 8 cm and 2 cm, then height of the object is
[KCET (Engg./Med.) 2000, 2001]
(a) 16 cm (b) 8 cm (c) 4 cm (d) 2 cm
Solution: (c) By using O = I1 I 2 ⇒ O = 8 × 2 = 4 cm
Example: 31
A convex lens produces a real image m times the size of the object. What will be the distance of the object
Solution: (a) from the lens
Example: 32 [JIPMER 2002]
Solution: (a) (a) m+ 1 f (b) (m − 1)f (c) m− 1 f (d) m+1
m m f
By using m = f here −m = (+ f ) ⇒ − 1 = f +u =1+ u ⇒u = − m+ 1 . f
+u (+ f ) + u m f f m
f
An air bubble in a glass sphere having 4 cm diameter appears 1 cm from surface nearest to eye when looked
along diameter. If a µ g = 1.5 , the distance of bubble from refracting surface is [CPMT 2002]
(a) 1.2 cm (b) 3.2 cm (c) 2.8 cm (d) 1.6 cm
By using
µ2 − µ1 = µ2 − µ1 µ2=1
v u R µ1=1.5
where u = ? , v = – 1 cm, µ1 = 1.5 , µ 2 = 1 , R = – 2 cm. C
1 − 1.5 = 1 − 1.5 ⇒ u = − 6 = −1.2 cm. u
−1 u 5
(− 2) v =1cm
R = 2cm
Example: 33 The sun's diameter is 1.4 × 109 m and its distance from the earth is 1011 m . The diameter of its image, formed
Solution: (c)
by a convex lens of focal length 2m will be [MP PET 2000]
Example: 34
Solution: (a) (a) 0.7 cm (b) 1.4 cm (c) 2.8 cm (d) Zero (i.e. point image)
From figure f
D 1011 ⇒ d= 2 × 1.4 × 109 = 2.8 cm. Sun α (d)
d= 2 1011 (D) α Image
1011 m
Two point light sources are 24 cm apart. Where should a convex lens of focal length 9 cm be put in between
them from one source so that the images of both the sources are formed at the same place
(a) 6 cm (b) 9 cm (c) 12 cm (d) 15 cm
The given condition will be satisfied only if one source (S1) placed on one side such that u < f (i.e. it lies
under the focus). The other source (S2) is placed on the other side of the lens such that u > f (i.e. it lies
beyond the focus).
If S1 is the object for lens then 1 = 1 − 1 1 11 ........(i)
f −y −x ⇒ y=x−f
If S2 is the object for lens then 1 = 1 − 1 ⇒ 1 = 11 ........(ii)
f +y − (24 − x) y f − (24 − x)
From equation (i) and (ii) I1 S1 S2
1 1 11 1 1 2 2 x2 I2
x −f f − (24 − x) x + (24 − x) f 9
= ⇒ = = ⇒ − 24 x + 108 = 0 x (24 – 4)
24 cm
On solving the equation x = 18 cm , 6 cm
y
Example: 35 There is an equiconvex glass lens with radius of each face as R and a µ g = 3 / 2 and a µw = 4 / 3 . If there is
Solution: (c) water in object space and air in image space, then the focal length is
(a) 2R (b) R (c) 3 R/2 (d) R 2
Consider the refraction of the first surface i.e. refraction from rarer medium to denser medium
3 − 4 43
2 3
µ2 − µ1 = µ1 + µ2 ⇒ = 3 + 2 ⇒ v1 = 9R
R −u v1 R ∞ v1
I I1
Now consider the refraction at the second surface of the lens i.e. Water Air
refraction from denser medium to rarer medium
1 − 3 3
2
= − 2 1 ⇒ v2 = 3 R
−R 9R + v2 2
The image will be formed at a distance do 3 R . This is equal to the focal length of the lens.
2
Tricky example: 4
A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm. On the
other side of the lens. At what distance from the lens a convex mirror of radius of curvature 10 cm be
placed in order to have an upright image of the object coincident with it
[CBSE PMT 1998; JIPMER 2001, 2002]
(a) 12 cm (b) 30 cm (c) 50 cm (d) 60 cm
Solution : (c) For lens u = 30 cm, f = 20 cm , hence by using 1 = 1 − 1 ⇒ 1 = 1 − 1 ⇒ v = 60 cm
f v u + 20 v − 30
The final image will coincide the object, if light ray falls normally O I
on convex mirror as shown. From figure it is seen clear that 30 cm 10 cm
reparation between lens and mirror is 60 – 10 = 50 cm. 60 cm
Tricky example: 5
A convex lens of local length 30 cm and a concave lens of 10 cm focal length are placed so as to have
the same axis. If a parallel beam of light falling on convex lens leaves concave lens as a parallel beam,
then the distance between two lenses will be
(a) 40 cm (b) 30 cm (c) 20 cm (d) 10 cm
d
Solution : (c) According to figure the combination behaves as plane glass plate (i.e., F= ∞)
Hence by using 111 d
F = f1 + f2 − f1 f2
⇒ 11 1 d ⇒ d = 20 cm
∞ = + 30 + − 10 − (30)(− 10)
Prism
Prism is a transparent medium bounded by refracting surfaces, such that the incident surface (on which light
ray is incidenting) and emergent surface (from which light rays emerges) are plane and non parallel.
Commonly used prism :
Equilateral prism Right angle prism Right angled isosceles prism
(1) Refraction through a prism
A A = r1 + r2 and i + e = A + δ
Aδ e
i r1 r2 For surface AC µ = sin i ; i – Angle of incidence, e – Angle of emergence,
sin r1 A – Angle of prism or refracting angle of prism,
r1 and r2 – Angle of refraction,
µ sin r2 δ – Angle of deviation
C sin e
B For surface AB µ =
(2) Deviation through a prism
For thin prism δ = (µ − 1)A . Also deviation is different for different colour light e.g. µ R < µV so δ R < δ V .
µFlint > µCrown so δ F > δ C
Maximum deviation Minimum deviation
δmax e It is observed if
i = 90o r1 = C r2
i rr ∠i = ∠e and ∠r1 = ∠r2 = r
e
then :
In this condition of maximum deviation ∠ i = 90o , r1 = C, (i) Refracted ray inside the prism is parallel to the base of
r2 = A − C and from Snell’s law on emergent surface
the prism δ
e = sin −1 sin(A − C)
sin C δm
i
(ii) r = A and i = A + δm
2 2
sin A +δm
2
(iii) µ = sin i or µ=
sin A / 2 sin A / 2
Note : ≅ If δ m = A then µ = 2 cos A / 2
(3) Normal incidence on a prism
If light ray incident normally on any surface of prism as shown
i = 0o e and e = 0o or
r1 = 0o r2 i r1 r2 = 0o
In any of the above case use µ = sin i and δ =i− A
sin A
(4) Grazing emergence and TIR through a prism
When a light ray falls on one surface of prism, it is not necessary that it will exit out from the prism. It may or
may not be exit out as shown below
Normal incidence Grazing incidence
Ray –1 : General emergence Ray –1 : General emergence
A < C and A < 2C and
A µ < cosec A A µ < cosec (A/2)
Ray – 2: Grazing emergence Ray – 2: Grazing emergence
A = C and
Ray - 3 Ray - 1 µ = cosec A Ray - 3 Ray - 1 A = 2C and
µ = cosec (A/2)
Ray – 3: TIR
Ray - 2 A > C and Ray - 2 Ray – 3: TIR
µ > cosec A A > 2C and
µ > cosec (A/2)
A = angle of prism and C = Critical angle for the material of
the prism
Note : ≅ For the condition of grazing emergence. Minimum angle of incidence
i min = sin −1 μ2 − 1 sinA − cosA .
(5) Dispersion through a prism
The splitting of white light into it’s constituent colours is called dispersion of light.
Incident Screen
white light
R
Y
V
(i) Angular dispersion (θ ) : Angular separation between extreme colours i.e. θ = δV − δR = (μV − μR )A . It
depends upon µ and A.
(ii) Dispersive power (ω) : ω= θ = µV − µR where µ y = µV + µ R
δy µy −1 2
⇒ It depends only upon the material of the prism i.e. µ and it doesn't depends upon angle of prism A
Note : ≅ Remember ω Flint > ω Crown .
(6) Combination of prisms
Two prisms (made of crown and flint material) are combined to get either dispersion only or deviation only.
Dispersion without deviation (chromatic combination) Deviation without dispersion (Achromatic combination)
Flint V Flint
A R
A
R A′ R A′
V
V
Crown Crown
(i) A' = − (µy − 1) (i) A' = − (µ V − µR)
A (µ'y −1) A (µ'V −µ 'R )
(ii) θ net = θ 1 − ω' = (ωδ − ω'δ ') (ii) δ net =δ 1 − ω
ω ω'
Scattering of Light
Molecules of a medium after absorbing incoming light radiations, emits them in all direction. This
phenomenon is called Scattering.
(1) According to scientist Rayleigh : Intensity of scattered light ∝ 1
λ4
(2) Some phenomenon based on scattering : (i) Sky looks blue due to scattering.
(ii) At the time of sunrise or sunset it looks reddish. (iii) Danger signals are made from red.
(3) Elastic scattering : When the wavelength of radiation remains unchanged, the scattering is called elastic.
(4) Inelastic scattering (Raman’s effect) : Under specific condition, light can also suffer inelastic scattering
from molecules in which it’s wavelength changes.
Rainbow
Rainbow is formed due to the dispersion of light suffering refraction and TIR in the droplets present in the
atmosphere.
(1) Primary rainbow : (i) Two refraction and one TIR. (ii) RedViolet
Innermost arc is violet and outermost is red. (iii) Subtends an
angle of 42o at the eye of the observer. (iv) More bright 42o 40o
(2) Secondary rainbow : (i) Two refraction and two TIR.
(ii) Innermost arc is red and outermost is violet.
(iii) It subtends an angle of 52.5o at the eye. (iv)
Comparatively less bright.
Colours
Colour is defined as the sensation received by the eye (rod cells of the eye) due to light coming from an object.
(1) Types of colours
Spectral colours Colours of pigment and dyes
Green (P) Yellow (P)
Cyan (S) Yellow (S) Green (S) Orange (S)
white Black
Blue (P) Magenta (S) Red (P) Blue (P) Red (P)
Radish violet (S)
(Mauve)
(i) Complementary colours : (i) Complementary colours :
Green and magenta yellow and mauve
Blue and yellow Red and green
Red and cyan Blue and orange
(ii) Combination : (ii) Combination :
Green + red + blue = White Yellow + red + blue = Black
Blue + yellow = White Blue + orange = Black
Red + cyan = White Red + green = Black
Green + magenta = White Yellow + mauve = Black
(2) Colours of object : The perception of a colour by eye depends on the nature of object and the light
incident on it.
Colours of opaque object Colours of transparent object
(i) Due to selective reflection. (i) Due to selective transmission.
(ii) A rose appears red in white light because it reflects (ii) A red glass appears red because it absorbs all
red colour and absorbs all remaining colours. colours, except red which it transmits.
(iii) When yellow light falls on a bunch of flowers, then (iii) When we look on objects through a green glass or
yellow and white flowers looks yellow. Other flowers green filter then green and white objects will appear
looks black. green while other black.
Note : ≅A hot object will emit light of that colour only which it has observed when it was heated.
Spectrum.
The ordered arrangements of radiations according to wavelengths or frequencies is called Spectrum. Spectrum
can be divided in two parts (I) Emission spectrum and (II) Absorption spectrum.
(1) Emission spectrum : When light emitted by a self luminous object is dispersed by a prism to get the
spectrum, the spectrum is called emission spectra.
Continuous emission Line emission spectrum Band emission spectrum
spectrum
(i) It consist of distinct bright lines. (iii) It consist of district bright bands.
(i) It consists of continuously varying
wavelengths in a definite wavelength (ii) It is produced by an excited (ii) It is produced by an excited
range. source in atomic state. source in molecular state.
(ii) It is produced by solids, liquids (iii) e.g. Spectrum of excited helium, (iii) e.g. Spectra of molecular H 2 ,
and highly compressed gases heated mercury vapours, sodium vapours or CO, NH 3 etc.
to high temperature. atomic hydrogen.
(iii) e.g. Light from the sun, filament
of incandescent bulb, candle flame
etc.
(2) Absorption spectrum : When white light passes through a semi-transparent solid, or liquid or gas, it’s
spectrum contains certain dark lines or bands, such spectrum is called absorption spectrum (of the substance
through which light is passed).
(i) Substances in atomic state produces line absorption spectra. Polyatomic substances such as H 2, CO2 and
KMnO4 produces band absorption spectrum.
(ii) Absorption spectra of sodium vapour have two (yellow lines) wavelengths D1(5890 Å) and D2(5896 Å)
Note : ≅ If a substance emits spectral lines at high temperature then it absorbs the same lines at low
temperature. This is Kirchoff’s law.
(3) Fraunhoffer’s lines : The central part (photosphere) of the sun is very hot and emits all possible
wavelengths of the visible light. However, the outer part (chromosphere) consists of vapours of different elements.
When the light emitted from the photosphere passes through the chromosphere, certain wavelengths are absorbed.
Hence, in the spectrum of sunlight a large number of dark lines are seen called Fraunhoffer lines.
(i) The prominent lines in the yellow part of the visible spectrum were labelled as D-lines, those in blue part as
F-lines and in red part as C-line.
(ii) From the study of Fraunhoffer’s lines the presence of various elements in the sun’s atmosphere can be
identified e.g. abundance of hydrogen and helium.
(4) Spectrometer : A spectrometer is used for obtaining pure spectrum of a source in laboratory and
calculation of µ of material of prism and µ of a transparent liquid.
It consists of three parts : Collimator which provides a parallel beam of light; Prism Table for holding the prism
and Telescope for observing the spectrum and making measurements on it.
The telescope is first set for parallel rays and then collimator is set for parallel rays. When prism is set in
minimum deviation position, the spectrum seen is pure spectrum. Angle of prism (A) and angle of minimum
deviation (δ m) are measured and µ of material of prism is calculated using prism formula. For µ of a transparent
liquid, we take a hollow prism with thin glass sides. Fill it with the liquid and measure (δ m) and A of liquid prism. µ
of liquid is calculated using prism formula.
(5) Direct vision spectroscope : It is an instrument used to observe pure spectrum. It produces dispersion
without deviation with the help of n crown prisms and (n − 1) flint prisms alternately arranged in a tabular structure.
For no deviation n(µ − 1)A = (n − 1) (µ '−1)A' .
Concepts
When a ray of white light passes through a glass prism red light is deviated less than blue light.
For a hollow prism A ≠ 0 but δ = 0
If an opaque coloured object or crystal is crushed to fine powder it will appear white (in sun light) as it will lose it's property of
selective reflection.
Our eye is most sensitive to that part at the spectrum which lies between the F line (sky green) one the C-line (red) of hydrogen
equal to the refractive index for the D line (yellow) of sodium. Hence for the dispersive power, the following formula is
internationally accepted ω = µF − µC
µD −1
Sometimes a part of prism is given and we keep on thinking whether how should we proceed ? To solve such problems first
complete the prism then solve as the problems of prism are solved A
50o
⇒
60o 70o 60o 70o
B C
Some other types of prism
Example
Example: 36 When light rays are incident on a prism at an angle of 45o, the minimum deviation is obtained. If refractive
Solution: (d) index of the material of prism is 2 , then the angle of prism will be [MP PMT 1986]
Example: 37
(a) 30o (b) 40o (c) 50o (d) 60o
Solution: (c)
Example: 38 1
Solution: (c)
Example: 39 µ = sin i ⇒ 2 = sin 45 ⇒ sin A = 2 = 1 ⇒ A = 30 o ⇒ A = 60o
Solution: (a) 2 2 2 2
Example: 40 sin A sin A
Solution: (c) 2 2
Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. The angle of
prism is (cos 41o = 0.75) [MP PET/PMT 1988]
(a) 62o (b) 41o (c) 82o (d) 31o
sin A +δm sin A + A
2 2
Given δm = A, then by using µ = ⇒µ = = sin A = 2 cos A sin A = 2 sin A cos A
2 2
sin A sin A sin A 2
2 2 2
⇒ 1.5 = 2 cos A ⇒ 0.75 = cos A ⇒ 41o = A ⇒ A = 82o .
2 2 2
Angle of glass prism is 60o and refractive index of the material of the prism is 1.414,then what will be the angle
of incidence, so that ray should pass symmetrically through prism
(a) 38o 61' (b) 35o 35' (c) 45o (d) 53o 8'
incident ray and emergent ray are symmetrical in the cure, when prism is in minimum deviation position.
Hence in this condition µ = sin i ⇒ sin i = µ sin A ⇒ sin i = 1.414 × sin 30o = 1 ⇒ i = 45o
2 2
sin A
2
A prism (µ = 1.5) has the refracting angle of 30o. The deviation of a monochromatic ray incident normally on
its one surface will be (sin 48 o 36' = 0.75) [MP PMT/PET 1988]
(a) 18o 36' (b) 20o 30' (c) 18o (d) 22o 1'
By using µ = sin i ⇒ 1.5 = sin i ⇒ sin i = 0.75 ⇒ i = 48 o 36'
sin A sin 30
Also from δ = i − A ⇒ δ = 48 o 36'−30o = 18 o 36'
Angle of a prism is 30o and its refractive index is 2 and one of the surface is silvered. At what angle of
incidence, a ray should be incident on one surface so that after reflection from the silvered surface, it retraces
its path [MP PMT 1991; UPSEAT 2001]
(a) 30o (b) 60o (c) 45o (d) sin −1 1.5
This is the case when light ray is falling normally an second surface.
Hence by using µ = sin i ⇒ 2 = sin i ⇒ sin i = 2 × 1 ⇒ i = 45 o
sin A sin 30o 2
Example: 41 The refracting angle of prism is A and refractive index of material of prism is cot A . The angle of minimum
Solution: (d) 2
Example: 42
Solution: (d) deviation is [CPMT 1992]
Example: 43
(a) 180o − 3A (b) 180o + 2A (c) 90o − A (d) 180o − 2A
Solution: (c)
Example: 44 sin A +δm sin A +δm cos A sin A +δm
2 2 sin 2 2
By using µ = A ⇒ cot A = A ⇒ A = A
2 2 2 2 2
sin sin sin
⇒ sin 90 − A = sin A +δm ⇒ 90 − A = A +δm ⇒ δm = 180 − 2A
2 2 2 2
A ray of light passes through an equilateral glass prism in such a manner that the angle of incidence is equal to
the angle of emergence and each of these angles is equal to 3/4 of the angle of the prism. The angle of
deviation is [MNR 1988; MP PMT 1999; Roorkee 2000; UPSEAT 2000]
(a) 45o (b) 39o (c) 20o (d) 30o
Given that A = 60o and i=e= 3 A= 3 × 60 = 45 o
4 4
By using i + e = A + δ ⇒ 45 + 45 = 60 + δ ⇒ δ = 30o
PQR is a right angled prism with other angles as 60o and 30o. Refractive index of prism is 1.5. PQ has a thin
layer of liquid. Light falls normally on the face PR. For total internal reflection, maximum refractive index of
liquid is
(a) 1.4 P 60° θ 30° Q
(b) 1.3
(c) 1.2
(d) 1.6 R
For TIR at PQ θ < C
From geometry of figure θ = 60 i.e. 60 > C ⇒ sin 60 > sin C
⇒ 3 > µ Liquid ⇒ µ Liquid < 3 × µ Pr ism ⇒ µ Liquid < 3 × 1.5 ⇒ µ Liquid < 1.3 .
2 µ Pr ism 2 2
Two identical prisms 1 and 2, each will angles of 30o, 60o and 90o are placed in contact as shown in figure. A
ray of light passed through the combination in the position of minimum deviation and suffers a deviation of
30o. If the prism 2 is removed, then the angle of deviation of the same ray is [PMT (Andhra) 1995]
(a) Equal to 15o 30o 30o
(b) Smaller than 30o 60o 90o 90o 60o
(c) More than 15o
Solution: (a) (d) Equal to 30o
δ = (µ − 1)A as A is halved, so δ is also halves
Example: 45 A prism having an apex angle 4o and refraction index 1.5 is located in front of a vertical plane mirror as shown
in figure. Through what total angle is the ray deviated after reflection from the mirror
(a) 176o 4°
(b) 4o 2°
(c) 178o
(d) 2o 2° 2°
Solution: (c) δ Pr ism = (µ − 1)A = (1.5 − 1)4 o = 2o
Example: 46
∴ δ Total = δ Pr ism + δ Mirror = (µ − 1)A + (180 − 2i) = 2o + (180 − 2 × 2) = 178 o
Solution: (d)
A ray of light is incident to the hypotenuse of a right-angled prism after travelling parallel to the base inside the
Example: 47 prism. If µ is the refractive index of the material of the prism, the maximum value of the base angle for which
Solution: (a)
Example: 48 light is totally reflected from the hypotenuse is [EAMCET 2003]
Solution: (a) (a) sin −1 1 (b) tan −1 1 (c) sin −1 µ − 1 (d) cos −1 1
µ µ µ µ
If α = maximum value of vase angle for which light is totally reflected from hypotenuse.
(90 − α) = C = minimum value of angle of incidence an hypotenuse for TIR
sin(90 − α) = sin C = 1 ⇒ α = cos −1 1 (90-α)
µ µ (90-α)
90° α
If the refractive indices of crown glass for red, yellow and violet colours are 1.5140, 1.5170 and 1.5318
respectively and for flint glass these are 1.6434, 1.6499 and 1.6852 respectively, then the dispersive powers
for crown and flint glass are respectively
[MP PET/PMT 1988]
(a) 0.034 and 0.064 (b) 0.064 and 0.034 (c) 1.00 and 0.064 (d) 0.034 and 1.0
ω Crown = µv − µr = 1.5318 − 1.5140 = 0.034 and ω Flint = µ ' − µ ' = 1.6852 − 1.6434 = 0.064
µy −1 (1.5170 − 1) v r 1.6499 − 1
µ ' −1
y
Flint glass prism is joined by a crown glass prism to produce dispersion without deviation. The refractive
indices of these for mean rays are 1.602 and 1.500 respectively. Angle of prism of flint prism is 10o, then the
angle of prism for crown prism will be [DPMT 2001]
(a) 12o 2.4' (b) 12o 4' (c) 1.24 o (d) 12o
For dispersion without deviation AC = (µ F − 1) ⇒ A = (1.602 − 1) ⇒ A = 12.04 o = 12o 2.4'
AF (µ C − 1) 10 (1.500 − 1)
Tricky example: 6
An achromatic prism is made by crown glass prism (AC = 19o ) and flint glass prism (AF = 6o ) . If
Cµv = 1.5 and Fµv = 1.66 , then resultant deviation for red coloured ray will be
(a) 1.04o (b) 5o (c) 0.96o (d) 13.5o
Solution : (d) For achromatic combination wC = −wF ⇒ [(µv − µr )A]C = −[(µv − µr )A]F
⇒ [µr A]C + [µr A]F = [µv A]C + [µv A]F = 1.5 × 19 + 6 × 1.66 = 38.5
Resultant deviation δ = [(µr − 1)A]C + [(µr − 1)A]F
= [µr A]C + [µr A]F − (AC + AF ) = 38.5 − (19 + 6) = 13.5o
Tricky example: 7
The light is incident at an angle of 60o on a prism of which the refracting angle of prism is 30o. The
refractive index of material of prism will be
(a) 2 (b) 2 3 (c) 2 (d) 3
Solution : (d) By using i + e = A + δ ⇒ 60 + e = 30 + 30 ⇒ e = 0 .
Hence ray will emerge out normally so by using the formula µ = sin i = sin 60 = 3
sin A sin 30
60°
Wave optics
Light Propagation.
Light is a form of energy which generally gives the sensation of sight.
(1) Different theories
Newtons Huygen’s Maxwell’s EM Einstein’s de-Broglie’s dual
corpuscular theory wave theory wave theory quantum theory theory of light
(i) Based on Rectilinear (i) Light propagates
propagation of light (i) Light travels in (i) Light travels in the (i) Light is produced, both as particles as
a hypothetical form of EM waves absorbed and well as waves
(ii) Light propagates in medium ether with speed in free propagated as
the form of tiny (high elasticity space c = 1 packets of energy (ii) Wave nature of
particles called very low density) called photons light dominates
Corpuscles. Colour of as waves µ0ε 0 when light interacts
light is due to different with light. The
size of corpuscles (ii) He proposed (ii) EM waves (ii) Energy associated particle nature of
that light waves consists of electric with each photon light dominates
are of longitudinal and magnetic field E = hν = hc when the light
nature. Later on it oscillation and they interacts with matter
was found that do not require λ (micro-scopic
they are material medium to h = planks constant particles )
transverse travel
= 6.6 × 10 −34 J - sec
ν = frequency
λ = wavelength
(2) Optical phenomena explained (√) or not explained (×) by the different theories of light
S. No. Phenomena Corpuscular Wave Theory Quantum Dual
√√ E.M. wave √ √
(i) Rectilinear Propagation √√ √ √
(ii) Reflection √ √ √
(iii) Refraction √√ √ √
(iv) Dispersion √ × √
(v) Interference ×√ √ × √
(vi) Diffraction ×√ √ × √
(vii) Polarisation ×√ × √
(viii) Double refraction ×√ √ × √
(ix) Doppler’s effect ×√ × √
(x) Photoelectric effect ×√ √ √
√
××
√
×
(3) Wave front
(i) Suggested by Huygens
(ii) The locus of all particles in a medium, vibrating in the same phase is called Wave Front (WF)
(iii) The direction of propagation of light (ray of light) is perpendicular to the WF.
(iv) Types of wave front.
Spherical WF Cylindrical WF Plane WF
Light ray Light rays
Point source Line source
(v) Every point on the given wave front acts as a source of new disturbance called secondary wavelets. Which
travel in all directions with the velocity of light in the medium.
A surface touching these secondary wavelets tangentially in the Point source
forward direction at any instant gives the new wave front at that instant.
This is called secondary wave front Secondary
wavelets
Primary Secondary
wave front wave front
Note : ≅Wave front always travels in the forward direction of the medium.
≅ Light rays is always normal to the wave front.
≅ The phase difference between various particles on the wave front is zero.
Principle of Super Position.
When two or more than two waves superimpose over each other at a common particle of the medium then
the resultant displacement (y) of the particle is equal to the vector sum of the displacements (y1 and y2) produced by
individual waves. i.e. y = y1 + y 2
(1) Graphical view :
(i) Resultant
1
Waves are meeting at a point
2 in same phase
y1 + y2 = y = y1 + y2
(ii) 1 Resultant
2
y = y1 – y2 Waves are meeting at a point
with out of phase
y1 + y2 =
(2) Phase / Phase difference / Path difference / Time difference
(i) Phase : The argument of sine or cosine in the expression for displacement of a wave is defined as the
phase. For displacement y = a sin ω t ; term ω t = phase or instantaneous phase
(ii) Phase difference (φ) : The difference between the phases of two waves at a point is called phase difference
i.e. if y1 = a1 sinω t and y2 = a2 sin (ω t + φ) so phase difference = φ
(iii) Path difference (∆) : The difference in path length’s of two waves meeting at a point is called path
difference between the waves at that point. Also ∆ = λ ×φ
2π
(iv) Time difference (T.D.) : Time difference between the waves meeting at a point is T.D. = T ×φ
2π
(3) Resultant amplitude and intensity
If suppose we have two waves y1 = a1 sinω t and y2 = a2 sin (ω t + φ) ; where a1, a2 = Individual amplitudes,
φ = Phase difference between the waves at an instant when they are meeting a point. I1, I2 = Intensities of
individual waves
Resultant amplitude : After superimposition of the given waves resultant amplitude (or the amplitude of
resultant wave) is given by A = a12 + a 2 + 2a1a 2 cosφ
2
For the interfering waves y1 = a1 sinω t and y2 = a2 cosω t, Phase difference between them is 90o. So resultant
amplitude A = a12 + a 2
2
Resultant intensity : As we know intensity ∝ (Amplitude)2 ⇒ I1 = ka12 , I 2 = ka 2 and I = kA2 (k is a
2
proportionality constant). Hence from the formula of resultant amplitude, we get the following formula of resultant
intensity I = I1 + I2 + 2 I1I2 cosφ
Note : ≅The term 2 I1I 2 cosφ is called interference term. For incoherent interference this term is zero so
resultant intensity I = I1 + I 2
(4) Coherent sources
The sources of light which emits continuous light waves of the same wavelength, same frequency and in same
phase or having a constant phase difference are called coherent sources.
Two coherent sources are produced from a single source of light by adopting any one of the following two methods
Division of wave front Division of amplitude
The light source is narrow
Light sources is extended. Light wave partly reflected
The wave front emitted by a narrow source is divided in (50%) and partly transmitted (50%)
two parts by reflection of refraction. The amplitude of wave emitted by an extend source of light is
The coherent sources obtained are imaginary e.g. Fresnel's divided in two parts by partial reflection and partial refraction.
biprism, Llyod's mirror Youngs' double slit etc.
The coherent sources obtained are real e.g. Newtons rings,
Michelson's interferrometer colours in thin films
S1 S Two waves M
L superimpose Reflection
S coating
S2
M2
Note : ≅Laser light is highly coherent and monochromatic.
≅ Two sources of light, whose frequencies are not same and phase difference between the waves
emitted by them does not remain constant w.r.t. time are called non-coherent.
≅ The light emitted by two independent sources (candles, bulbs etc.) is non-coherent and interference
phenomenon cannot be produced by such two sources.
≅ The average time interval in which a photon or a wave packet is emitted from an atom is defined as
the time of coherence. It is τc = L = Distance of coherence , it's value is of the order of 10–10 sec.
c Velocity of light
Interference of Light.
When two waves of exactly same frequency (coming from two coherent sources) travels in a medium, in the
same direction simultaneously then due to their superposition, at some points intensity of light is maximum while at
some other points intensity is minimum. This phenomenon is called Interference of light.
(1) Types : It is of following two types
Constructive interference Destructive interference
(i) When the waves meets a point with same phase, (i) When the wave meets a point with opposite phase,
constructive interference is obtained at that point (i.e. destructive interference is obtained at that point (i.e.
maximum light) minimum light)
(ii) Phase difference between the waves at the point of (ii) φ = 180o or (2n − 1)π ; n = 1, 2, ...
observation φ = 0o or 2nπ
or (2n + 1)π ; n = 0,1,2.....
(iii) Path difference between the waves at the point of (iii) ∆ = (2n − 1) λ (i.e. odd multiple of λ/2)
2
observation ∆ = nλ (i.e. even multiple of λ/2)
(iv) Resultant amplitude at the point of observation will
(iv) Resultant amplitude at the point of observation will be be minimum
maximum
Amin = a1 − a2
a1 = a2 ⇒ Amin = 0
If a1 = a2 = a0 ⇒ Amax = 2a0 If a1 = a2 ⇒ Amin = 0
(v) Resultant intensity at the point of observation will be (v) Resultant intensity at the point of observation will be
maximum minimum
I max = I1 + I 2 + 2 I1 I 2 I min = I1 + I 2 − 2 I1 I 2
( )I max = I1 + I 2 2 ( )I min = I1 − I 2 2
If I1 = I 2 = I 0 ⇒ I max = 2I 0 If I1 = I 2 = I 0 ⇒ I min = 0
(2) Resultant intensity due to two identical waves :
For two coherent sources the resultant intensity is given by I = I1 + I 2 + 2 I1I 2 cosφ
For identical source I1 = I 2 = I0 ⇒ I = I0 + I0 + 2 I0 I0 cosφ = 4I0 cos 2 φ [1 + cosθ = 2 cos 2 θ ]
2 2
Note : ≅ In interference redistribution of energy takes place in the form of maxima and minima.
≅ Average intensity : I av = I max + I min = I1 + I2 = a12 + a 2
2 2
≅ Ratio of maximum and minimum intensities :
I max I1 + I2 2 I1 / I2 + 1 2 a1 + a2 2 a1 / a2 1 2 I1 a1 I max + 1
I min I1 − I2 I1 / I2 − 1 a1 − a2 a1 / a2 1 I2 a2 I min
= = = = + also = =
− I max
I min − 1
≅ If two waves having equal intensity (I1 = I2 = I0) meets at two locations P and Q with path difference
∆1 and ∆2 respectively then the ratio of resultant intensity at point P and Q will be
IP = cos 2 φ1 cos 2 π∆1
IQ cos 2 2 = λ
φ2
2 cos 2 π∆ 2
λ
Young’s Double Slit Experiment (YDSE)
Monochromatic light (single wavelength) falls on two narrow slits S1 and S2 which are very close together acts
as two coherent sources, when waves coming from two coherent sources (S1, S 2 ) superimposes on each other, an
interference pattern is obtained on the screen. In YDSE alternate bright and dark bands obtained on the screen.
These bands are called Fringes.
S S1 3 Bright Screen Central bright fringe
d 2 Bright 4 Dark (or Central maxima)
d = Distance between slits 1 Bright 3 Dark
S2 2 Dark
D = Distance between slits and screen 1 Bright 1 Dark
λ = Wavelength of monochromatic light 2 Bright 1 Dark
3 Bright 2 Dark
emitted from source D 3 Dark
4 Dark
(1) Central fringe is always bright, because at central position φ = 0o or ∆ = 0
(2) The fringe pattern obtained due to a slit is more bright than that due to a point.
(3) If the slit widths are unequal, the minima will not be complete dark. For very large width uniform
illumination occurs.
(4) If one slit is illuminated with red light and the other slit is illuminated with blue light, no interference pattern
is observed on the screen.
(5) If the two coherent sources consist of object and it’s reflected image, the central fringe is dark instead of
bright one.
(6) Path difference P
x
Path difference between the interfering waves meeting at a point P on the screen C
is given by ∆ = xd = d sinθ S1 Screen
D d
θ θ
M
where x is the position of point P from central maxima. S2
D
For maxima at P : ∆ = nλ ; where n = 0, ± 1, ± 2, …….
and For minima at P : ∆ = (2n − 1)λ ; where n = ± 1, ± 2, …….
2
Note : ≅If the slits are vertical, the path difference (∆) is d sinθ , so as θ increases, ∆ also increases. But if slits
are horizontal path difference is d cosθ , so as θ increases, ∆ decreases.
PP
S1 C θ C
dθ S1 S2
S2 d
D
(7) More about fringe
(i) All fringes are of
equal width. Width of each fringe is β = λD and angular fringe width θ = λ = β
d d D
(ii) If the whole YDSE set up is taken in another medium then λ changes so β changes
e.g. in water λw = λa ⇒ βw = βa = 3 β a
µw µw 4
(iii) Fringe width β ∝ 1 i.e. with increase in separation between the sources, β decreases.
d
(iv) Position of nth bright fringe from central maxima xn = nλD = nβ ; n = 0,1, 2....
d
(v) Position of nth dark fringe from central maxima xn = (2n − 1) λD = (2n − 1) β ; n = 1, 2,3....
2d 2
(vi) In YDSE, if n1 fringes are visible in a field of view with light of wavelength λ1 , while n2 with light of
wavelength λ2 in the same field, then n1λ1 = n2λ2 .
(vii) Separation (∆x) between fringes
Between nth bright and mth bright fringes (n > m) Between nth bright and mth dark fringe
∆x = (n − m)β (a) If n>m then ∆x = n − m + 1 β
2
(b) If n < m then ∆x = m − n − 1 β
2
(8) Identification of central bright fringe
To identify central bright fringe, monochromatic light is replaced by white light. Due to overlapping central maxima
will be white with red edges. On the other side of it we shall get a few coloured band and then uniform illumination.
(9) Condition for observing sustained interference
(i) The initial phase difference between the interfering waves must remain constant : Otherwise the interference
will not be sustained.
(ii) The frequency and wavelengths of two waves should be equal : If not the phase difference will not remain
constant and so the interference will not be sustained.
(iii) The light must be monochromatic : This eliminates overlapping of patterns as each wavelength
corresponds to one interference pattern. t
(iv) The amplitudes of the waves must be equal : This improves contrast
with I max = 4I 0 and I min = 0. µ
(v) The sources must be close to each other : Otherwise due to small fringe S1 C
d
width β ∝ 1 the eye can not resolve fringes resulting in uniform illumination. S2
d
(10) Shifting of fringe pattern in YDSE Screen
D
If a transparent thin film of mica or glass is put in the path of one of the
waves, then the whole fringe pattern gets shifted.
If film is put in the path of upper wave, fringe pattern shifts upward and if film is placed in the path of lower
wave, pattern shift downward.
Fringe shift = D (µ − 1)t = β (µ − 1)t
d λ
⇒ Additional path difference = (µ − 1)t
⇒ If shift is equivalent to n fringes then n = (µ − 1)t or t = nλ
λ (µ − 1)
⇒ Shift is independent of the order of fringe (i.e. shift of zero order maxima = shift of nth order maxima.
⇒ Shift is independent of wavelength.
(11) Fringe visibility (V)
With the help of visibility, knowledge about coherence, fringe contrast an interference pattern is obtained.
V = I max − I min = 2 I1 I 2 If I min = 0, V = 1 (maximum) i.e., fringe visibility will be best.
I max + I min (I1 + I 2 )
Also if I max = 0, V = −1 and If I max = I min , V = 0
(12) Missing wavelength in front of one of the slits in YDSE
From figure S2P = D 2 + d 2 and S1P = D S1 P
d Central
So the path difference between the waves reaching at P
S2 position
∆ = S2 P − S1P = D2 + d2 −D = D 1 + d2 1/ 2 −D
D2 D
From binomial expansion ∆ = D 1 + 1 d2 − D = d2
2 D2 2D
For Dark at P ∆ = d2 = (2n − 1)λ ⇒ Missing wavelength at P λ = d2
2D 2 (2n − 1)D
By putting n = 1, 2, 3.... Missing wavelengths are λ = d2 , d2 , d2 ....
D 3D 5D
Illustrations of Interference
Interference effects are commonly observed in thin films when their thickness is comparable to wavelength of
incident light (If it is too thin as compared to wavelength of light it appears dark and if it is too thick, this will result in
uniform illumination of film). Thin layer of oil on water surface and soap bubbles shows various colours in white
light due to interference of waves reflected from the two surfaces of the film.
Air Air
Oil
Air
Water Soap bubble in air
Oil film on water surface
(1) Thin films : In thin films interference takes place between the waves reflected from it’s two surfaces and
waves refracted through it.
Reflected rays
t µ r
r
Refracted rays
Interference in reflected light Interference in refracted light
Condition of constructive interference (maximum intensity) Condition of constructive interference (maximum intensity)
∆ = 2µ t cos r = (2n ± 1) λ ∆ = 2µ t cos r = (2n) λ
2 2
For normal incidence r = 0 For normal incidence
so 2µ t = (2n ± 1) λ2 2µ t = nλ
Condition of destructive interference (minimum intensity)
Condition of destructive interference (minimum intensity)
∆ = 2µ t cos r = (2n) λ ∆ = 2µ t cos r = (2n ± 1) λ
2 2
For normal incidence 2µ t = nλ For normal incidence 2µ t = (2n ± 1) λ
2
Note : ≅The Thickness of the film for interference in visible light is of the order of 10,000Å .
(2) Lloyd's Mirror
A plane glass plate (acting as a mirror) is illuminated at almost grazing incidence by a light from a slit S1. A
virtual image S2 of S1 is formed closed to S1 by reflection and these two act as coherent sources. The expression
giving the fringe width is the same as for the double slit, but the fringe system differs in one important respect.
In Lloyd's mirror, if the point P, for example, is such that the path difference S2 P − S1P is a whole number of
wavelengths, the fringe at P is dark not bright. This is due to 180o phase change which occurs when light is reflected
from a denser medium. This is equivalent to adding an extra half wavelength to the path of the reflected wave. At
grazing incidence a fringe is formed at O, where the geometrical path difference between the direct and reflected
waves is zero and it follows that it will be dark rather than bright.
P
S1
dO
S2
Thus, whenever there exists a phase difference of a π between the two interfering beams of light, conditions of
maximas and minimas are interchanged, i.e., ∆x = nλ (for minimum intensity)
and ∆x = (2n − 1)λ / 2 (for maximum intensity)
Doppler’s Effect in Light
The phenomenon of apparent change in frequency (or wavelength) of the light due to relative motion between
the source of light and the observer is called Doppler’s effect.
If ν = actual frequency, ν ' = Apparent frequency, v = speed of source w.r.t stationary observer, c = speed of light
Source of light moves towards the stationary Source of light moves away from the stationary
observer (v << c) observer (v << c)
(i) Apparent frequency ν′ =ν 1 + v and (i) Apparent frequency ν ′ = ν 1 − v and
c c
Apparent wavelength λ′ = λ 1 − v Apparent wavelength λ′ = λ 1 + v
c c
(ii) Doppler’s shift : Apparent wavelength < actual wavelength, (ii) Doppler’s shift : Apparent wavelength > actual wavelength,
So spectrum of the radiation from the source of light shifts So spectrum of the radiation from the source of light shifts
towards the red end of spectrum. This is called Red shift towards the violet end of spectrum. This is called Violet shift
Doppler’s shift Δλ = λ. v Doppler’s shift Δλ = λ. v
c c
Note : ≅Doppler’s shift (∆λ) and time period of rotation (T) of a star relates as ∆λ = λ × 2πr ; r = radius of star.
c T
Applications of Doppler effect
(i) Determination of speed of moving bodies (aeroplane, submarine etc) in RADAR and SONAR.
(ii) Determination of the velocities of stars and galaxies by spectral shift.
(iii) Determination of rotational motion of sun.
(iv) Explanation of width of spectral lines.
(v) Tracking of satellites. (vi) In medical sciences in echo cardiogram, sonography etc.
Concepts
The angular thickness of fringe width is defined as δ = β = λ , which is independent of the screen distance D.
D d
Central maxima means the maxima formed with zero optical path difference. It may be formed anywhere on the screen.
All the wavelengths produce their central maxima at the same position.
The wave with smaller wavelength from its maxima before the wave with longer wavelength.
The first maxima of violet colour is closest and that for the red colour is farthest.
Fringes with blue light are thicker than those for red light.
In an interference pattern, whatever energy disappears at the minimum, appears at the maximum.
In YDSE, the nth maxima always comes before the nth minima.
In YDSE, the ratio Imax is maximum when both the sources have same intensity.
Imin
For two interfering waves if initial phase difference between them is φ0 and phase difference due to path difference between them
2π
is φ '. Then total phase difference will be φ = φ0 +φ' = φ0 + λ ∆.
Sometimes maximm number of maximas or minimas are asked in the question which can be obtained on the screen. For this we
use the fact that value of sin θ (or cos θ) can't be greater than 1. For example in the first case when the slits are vertical
sin θ = nλ (for maximum intensity)
d
sin θ ≯1 ∴ nλ ≯ 1 or n ≯ d
d λ
Suppose in some question d/λ comes out say 4.6, then total number of maximuas on the screen will be 9. Corresponding to
n = 0, ± 1, ± 2, ± 3 and ± 4 .
Shape of wave front
If rays are parallel, wave front is plane. If rays are converging wave front is spherical of decreasing radius. If rays are diverging
wave front is spherical of increasing radius.
Wave front
Reflection and refraction of wave front
BD B C
ii r
ii rr A
r
D
AC
Reflection Refraction
BC = AD and ∠ i = ∠ r
BC = v1 = sin i = µ2
AD v2 sin r µ1
Example
Example: 1 If two light waves having same frequency have intensity ratio 4 : 1 and they interfere, the ratio of maximum to
Solution: (a) minimum intensity in the pattern will be [BHU 1995; MP PMT 1995; DPMT 1999; CPMT 2003]
Example: 2
Solution: (d) (a) 9 : 1 (b) 3 : 1 (c) 25 : 9 (d) 16 : 25
Example: 3
I max I1 + 1 2 4 + 1 2
Solution: (b) I min I2 1 9
By using = = 4 = 1 .
Example: 4 I1 1
I2 − 1
− 1
In Young’s double slit experiment using sodium light (λ = 5898Å), 92 fringes are seen. If given colour
(λ = 5461Å) is used, how many fringes will be seen [RPET 1996; JIPMER 2001, 2002]
(a) 62 (b) 67 (c) 85 (d) 99
By using n1λ1 = n2λ2 ⇒ 92 × 5898 = n2 × 5461 ⇒ n2 = 99
Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase
difference between the beams is π at point A and π at point B. Then the difference between the resultant
2
intensities at A and B is [IIT-JEE (Screening) 2001]
(a) 2I (b) 4I (c) 5I (d) 7I
By using I = I1 + I 2 + 2 I1 I 2 cos φ
At point A : Resultant intensity I A = I + 4I + 2 I × 4I cos π = 5I
2
At point B : Resultant intensity I B = I + 4I + 2 I × 4I cos π = I . Hence the difference = I A − I B = 4I
If two waves represented by y1 = 4 sin ωt and y2 = 3 sinωt + π interfere at a point, the amplitude of the resulting
3
wave will be about [MP PMT 2000]
(a) 7 (b) 6 (c) 5 (d) 3.
Solution: (b) By using A = a12 + a 2 + 2a1 a 2 cos φ ⇒ A= (4)2 + (3)2 + 2 × 4 × 3 cos π = 37 ≈ 6 .
Example: 5 2 3
Solution: (d) Two waves being produced by two sources S1 and S2 . Both sources have zero phase difference and have
Example: 6 wavelength λ. The destructive interference of both the waves will occur of point P if (S1P − S2P) has the value
Solution: (d)
Example: 7 [MP PET 1987]
Solution: (b) (a) 5λ (b) 3 λ (c) 2λ (d) 11 λ
Example: 8 4 2
Solution: (c) For destructive interference, path difference the waves meeting at P (i.e. S1P − S2 P) must be odd multiple of
Example: 9 λ/2. Hence option (d) is correct.
Solution: (b) Two interfering wave (having intensities are 9I and 4I) path difference between them is 11 λ. The resultant
intensity at this point will be
(a) I (b) 9 I (c) 4 I (d) 25 I
Path difference ∆ = λ ×φ ⇒ 2π × 11λ = 22π i.e. constructive interference obtained at the same point
2π λ
So, resultant intensity I R = ( I1 + I 2 )2 = ( 9I + 4I )2 = 25I .
In interference if I max = 144 then what will be the ratio of amplitudes of the interfering wave
I min 81
(a) 144 (b) 7 (c) 1 (d) 12
81 1 7 9
a1 I max + 1 144 + 1 12 + 1
a2 I min − 81 9 7
By using = = 144 = 12 = 1
I max 81 5
I min 1 − 1 − 1
Two interfering waves having intensities x and y meets a point with time difference 3T/2. What will be the
resultant intensity at that point
(a) ( x + y) (b) ( x + y + xy) (c) x + y + 2 xy (d) x+y
2xy
Time difference T.D. = T ×φ ⇒ 3T = T ×φ ⇒ φ = 3π ; This is the condition of constructive interference.
2π 2 2π
So resultant intensity I R = ( I1 + I 2 )2 = ( x + y)2 = x + y + 2 xy.
In Young’s double-slit experiment, an interference pattern is obtained on a screen by a light of wavelength
6000 Å, coming from the coherent sources S1 and S2 . At certain point P on the screen third dark fringe is
formed. Then the path difference S1P − S2P in microns is [EAMCET 2003]
(a) 0.75 (b) 1.5 (c) 3.0 (d) 4.5
For dark fringe path difference ∆ = (2n − 1) λ ; here n = 3 and λ = 6000 × 10–10 m
2
So ∆ = (2 × 3 − 1) × 6 × 10 −7 = 15 × 10 −7 m = 1.5 microns.
2
Example: 10 In a Young’s double slit experiment, the slit separation is 1 mm and the screen is 1 m from the slit. For a monochromatic
Solution: (b)
Example: 11 light of wavelength 500 nm, the distance of 3rd minima from the central maxima is [Orissa JEE 2003]
Solution: (b)
Example: 12 (a) 0.50 mm (b) 1.25 mm (c) 1.50 mm (d) 1.75 mm
Solution: (b)
Distance of nth minima from central maxima is given as x = (2n − 1)λ D
Example: 13 2d
Solution: (a)
Example: 14 So here x = (2 × 3 − 1) × 500 × 10 −9 ×1 = 1.25 mm
2 × 10 −3
Solution: (c)
Example: 15 The two slits at a distance of 1 mm are illuminated by the light of wavelength 6.5 ×10−7 m. The interference fringes are
observed on a screen placed at a distance of 1 m. The distance between third dark fringe and fifth bright fringe will be
[NCERT 1982; MP PET 1995; BVP 2003]
(a) 0.65 mm (b) 1.63 mm (c) 3.25 mm (d) 4.88 mm
Distance between nth bright and mth dark fringe (n > m) is given as x = n − m + 1 β = n − m + 1 λD
2 2 d
⇒ x = 5 − 3 + 1 × 6.5 × 10 −7 × 1 = 1.63 mm .
2 1 × 10 −3
The slits in a Young’s double slit experiment have equal widths and the source is placed symmetrically relative to the slits.
The intensity at the central fringes is I0. If one of the slits is closed, the intensity at this point will be [MP PMT 1999]
(a) I0 (b) I 0 / 4 (c) I 0 / 2 (d) 4I 0
By using IR = 4I cos 2 φ {where I = Intensity of each wave}
2
At central position φ = 0o, hence initially I0 = 4I.
If one slit is closed, no interference takes place so intensity at the same location will be I only i.e. intensity
become s 1 th or I0 .
4 4
In double slit experiment, the angular width of the fringes is 0.20° for the sodium light (λ = 5890 Å). In order to increase
the angular width of the fringes by 10%, the necessary change in the wavelength is [MP PMT 1997]
(a) Increase of 589 Å (b) Decrease of 589 Å (c) Increase of 6479 Å (d) Zero
By using θ λ ⇒ θ1 = λ1 0.20 o 5890 0.20 5890 λ2 = 6479
= d θ2 λ2 ⇒ (0.20o + 10% of 0.20) = λ2 ⇒ 0.22 = ⇒
λ2
So increase in wavelength = 6479 – 5890 = 589 Å.
In Young’s experiment, light of wavelength 4000 Å is used, and fringes are formed at 2 metre distance and has a fringe
width of 0.6 mm. If whole of the experiment is performed in a liquid of refractive index 1.5, then width of fringe will be
(a) 0.2 mm (b) 0.3 mm (c) 0.4 mm [MP PMT 1994, 97]
(d) 1.2 mm
β medium = β air ⇒ β medium = 0.6 = 0.4mm .
µ 1.5
Two identical sources emitted waves which produces intensity of k unit at a point on screen where path
difference is λ. What will be intensity at a point on screen at which path difference is λ/4 [RPET 1996]
(a) k (b) k (c) k (d) Zero
4 2
Solution: (b) By using phase difference φ = 2π (∆)
λ
Example: 16
Solution: (a) For path difference λ, phase difference φ1 = 2π and for path difference λ/4, phase difference φ2 = π/2.
Example: 17
Solution: (a) Also by using I = 4I0 cos 2 φ ⇒ I1 = cos 2 (φ1 / 2) ⇒ k = cos 2 (2π / 2) = 1 ⇒ I2 = k .
2 I2 cos 2 (φ 2 / 2) I2 1/ 2 2
Example: 18 cos 2 π / 2
Solution: (a) 2
Example: 19
A thin mica sheet of thickness 2 × 10−6 m and refractive index (µ = 1.5) is introduced in the path of the first
wave. The wavelength of the wave used is 5000Å. The central bright maximum will shift [CPMT 1999]
(a) 2 fringes upward (b) 2 fringes downward (c) 10 fringes upward (d) None of these
By using shift ∆x = p (µ − 1) t ⇒ ∆x = β (1.5 − 1) × 2 × 10 −6 = 2β
λ 5000 × 10 −10
Since the sheet is placed in the path of the first wave, so shift will be 2 fringes upward.
In a YDSE fringes are observed by using light of wavelength 4800 Å, if a glass plate (µ = 1.5) is introduced in
the path of one of the wave and another plates is introduced in the path of the (µ = 1.8) other wave. The
central fringe takes the position of fifth bright fringe. The thickness of plate will be
(a) 8 micron (b) 80 micron (c) 0.8 micron (d) None of these
Shift due to the first plate x1 = β (µ1 − 1) t (Upward)
λ
1
and shift due to the second x2 = β (µ 2 − 1) t (Downward) S1 2 C
λ d Screen
D
Hence net shift = x2 – x1 = β (µ 2 − µ1 ) t S2
λ
⇒ 5p = β (1.8 − 1.5) t ⇒ t = 5λ = 5 × 4800 × 10 −10 = 8 × 10 −6 m = 8 micron .
λ 0.3 0.3
In young double slit experiment d = 10 −4 (d = distance between slits, D = distance of screen from the slits).
D
At a point P on the screen resulting intensity is equal to the intensity due to individual slit I0. Then the distance
of point P from the central maxima is (λ = 6000 Å)
(a) 2 mm (b) 1 mm (c) 0.5 mm (d) 4 mm
By using shift I = 4I 0 cos 2 (φ / 2) ⇒ I 0 = 4I 0 cos 2 (φ / 2) ⇒ cos(φ / 2) = 1 or φ = π ⇒ φ = 2π
2 2 3 3
Also path difference ∆ = xd = λ ×φ ⇒ x × d = 6000 × 10 −10 × 2π ⇒ x = 2 × 10 −3 m = 2mm.
D 2π D 2π 3
Two identical radiators have a separation of d = λ/4, where λ is the wavelength of the waves emitted by either
source. The initial phase difference between the sources is π/4. Then the intensity on the screen at a distance
point situated at an angle θ = 30o from the radiators is (here I0 is the intensity at that point due to one
radiator)
(a) I0 (b) 2I0 (c) 3I0 (d) 4I0
Solution: (a) Initial phase difference φ0 = π ; Phase difference due to path difference φ ' = 2π (∆)
4 λ
Example: 20
Solution: (a) where ∆ = d sinθ ⇒ φ '= 2π (d sinθ ) = 2π × λ (sin 30o ) = π
Example: 21 λ λ 4 4
Solution: (a)
Hence total phase difference φ = φ0 +φ '= φ . By using I = 4I0 cos 2 (φ / 2) = 4I0 cos 2 π / 2 = 2I0 .
Example: 22 4 2
Solution: (b)
Example: 23 In YDSE a source of wavelength 6000 Å is used. The screen is placed 1 m from the slits. Fringes formed on
the screen, are observed by a student sitting close to the slits. The student's eye can distinguish two
neighbouring fringes. If they subtend an angle more than 1 minute of arc. What will be the maximum distance
between the slits so that the fringes are clearly visible
(a) 2.06 mm (b) 2.06 cm (c) 2.06 × 10–3 mm (d) None of these
According to given problem angular fringe width θ = λ ≥ π [As 1' = π 60 rad]
d 180 × 60 180 ×
i.e. d < 6 × 10 −7 × 180 × 60 i.e. d < 2.06 × 10 −3 m ⇒ dmax = 2.06 mm
π
the maximum intensity in case of interference of n identical waves, each of intensity I0, if the interference is (i)
coherent and (ii) incoherent respectively are
(a) n2 I 0 , nI 0 (b) nI 0 , n2 I 0 (c) nI 0 , I 0 (d) n2 I 0 ,(n − 1)I 0
In case of interference of two wave I = I1 + I 2 + 2 I1I 2 cos φ
(i) In case of coherent interference φ does not vary with time and so I will be maximum when cos φ = max = 1
i.e. (I max )co = I1 + I 2 + 2 I1 I 2 = ( I1 + I 2 )2
So for n identical waves each of intensity I0 (I max )co = ( I 0 + I 0 + ......)2 = (n I 0 )2 = n2 I 0
(ii)In case of incoherent interference at a given point, φ varies randomly with time, so (cos φ)av = 0 and hence
(I R )Inco = I1 + I 2
So in case of n identical waves (I R )Inco = I 0 + I 0 + ....... = nI 0
The width of one of the two slits in a Young's double slit experiment is double of the other slit. Assuming that
the amplitude of the light coming from a slit is proportional to the slit width. The ratio of the maximum to the
minimum intensity in interference pattern will be
(a) 1 (b) 9 (c) 2 (d) 1
a 1 1 2
I max Amax 2 3A 2 9
I min Amin A 1
Amax = 2A + A = 3A and Amin = 2A − A = A . Also = = =
A star is moving towards the earth with a speed of 4.5 ×106 m / s . If the true wavelength of a certain line in the
spectrum received from the star is 5890 Å, its apparent wavelength will be about [c = 3 ×108 m / s]
[MP PMT 1999]
(a) 5890 Å (b) 5978 Å (c) 5802 Å (d) 5896 Å
Solution: (c) By using λ' = λ 1 − v ⇒ λ' = 5890 1 − 4.5 × 106 = 5802 Å .
Example: 24 c 3 × 108
Solution: (b) Light coming from a star is observed to have a wavelength of 3737 Å, while its real wavelength is 3700 Å. The
Example: 25
speed of the star relative to the earth is [Speed of light = 3 ×108 m / s ] [MP PET 1997]
Solution: (a)
(a) 3 ×105 m / s (b) 3 ×106 m / s (c) 3.7 ×107 m / s (d) 3.7 ×106 m / s
By using ∆λ = λ v ⇒ (3737-3700)= 3700 × 3 v ⇒ v = 3 × 106 m / s .
c × 108
Light from the constellation Virgo is observed to increase in wavelength by 0.4%. With respect to Earth the
constellation is [MP PMT 1994, 97; MP PET 2003]
(a) Moving away with velocity 1.2 ×106m / s (b) Coming closer with velocity 1.2 ×106m / s
(c) Moving away with velocity 4 ×106m / s (d) Coming closer with velocity 4 ×106m / s
By using ∆λ = v ; where ∆λ = 0.4 and c = 3 × 108 m/s ⇒ 0.4 v ⇒ v = 1.2 × 106 m/s
λ c λ 100 100 = 3 × 108
Since wavelength is increasing i.e. it is moving away.
Tricky example: 1
In YDSE, distance between the slits is 2 × 10–3 m, slits are illuminated by a light of wavelength 2000Å –
9000 Å. In the field of view at a distance of 10–3 m from the central position which wavelength will be
observe. Given distance between slits and screen is 2.5 m
(a) 40000 Å (b) 4500 Å (c) 5000 Å (d) 5500 Å
Solution : (b) x= nλ D ⇒ λ = xd = 10−3 × 2 × 10−3 ⇒ 8 × 10 −7 m= 8000 Å
d nD n × 2.5 n n
For n = 1, 2, 3....... λ = 8000 Å, 4000 Å, 8000 Å .........
3
Hence only option (a) is correct.
Tricky example: 2
I is the intensity due to a source of light at any point P on the screen. If light reaches the point P via two
different paths (a) direct (b) after reflection from a plane mirror then path difference between two paths
is 3λ/2, the intensity at P is
(a) I (b) Zero (c) 2I (d) 4I
Solution : (d) Reflection of light from plane mirror gives additional path difference of λ/2 between two waves
∴ Total path difference = 3λ + λ = 2λ
2 2
Which satisfies the condition of maxima. Resultant intensity = ( I + I )2 = 4I.
Tricky example: 3
I B B′
A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in figure. It
undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The
rays AB and A′B′ undergo interference. The ratio Imax / Imin is [IIT-JEE 1990]
(a) 4 : 1
(b) 8 : 1
(c) 7 : 1
(d) 49 : 1
Solution : (d) From figure I1 = I and I2 = 9I ⇒ I2 9 B B′
4 64 I1 = 16 I /4 9I /64
I
I2 + 1 9 + 1 A A′ 3I /64
I1 − 1 16
By using I max = 9 = 49 3I /4 3I /16
I min = I2 16 1
I1 − 1
Fresnel's Biprism.
(1) It is an optical device of producing interference of light Fresnel's biprism is made by joining base to base
two thin prism (A1BC and A2BC as shown in the figure) of very small angle or by grinding a thick glass plate.
(2) Acute angle of prism is about 1/2o and obtuse angle of prism is about 179o.
(3) When a monochromatic light source is kept in front of biprism two coherent virtual source S1 and S2 are produced.
(4) Interference fringes are found on the screen (in the MN region) placed behind the biprism interference
fringes are formed in the limited region which can be observed with the help eye piece.
(5) Fringe width is measured by a micrometer attached to the eye piece. Fringes are of equal width and its
value is β = λD ⇒ λ = βd
d D
S1 A1 M
dS N
α
S2
CB
α b
D
A2
a
Let the separation between S1 and S2 be d and the distance of slits and the screen from the biprism be a and b
respectively i.e. D = (a + b). If angle of prism is α and refractive index is µ then d = 2a(µ − 1)α
∴ λ = β [2a (µ − 1)α] ⇒ β = (a + b)λ
(a + b) 2a(µ − 1)α
Diffraction of Light.
It is the phenomenon of bending of light around the corners of an obstacle/aperture of the size of the
wavelength of light.
Shadow Light
Light Shadow
Aperture Obstacle
Note : ≅Diffraction is the characteristic of all types of waves.
≅ Greater the wavelength of wave, higher will be it’s degree of diffraction.
≅ Experimental study of diffraction was extended by Newton as well as Young. Most systematic study
carried out by Huygens on the basis of wave theory.
≅The minimum distance at which the observer should be from the obstacle to observe the diffraction of
light of wavelength λ around the obstacle of size d is given by x = d2 .
4λ
(1) Types of diffraction : The diffraction phenomenon is divided into two types
Fresnel diffraction Fraunhofer diffraction
(i) If either source or screen or both are at finite (i) In this case both source and screen are
distance from the diffracting device (obstacle or effectively at infinite distance from the diffracting
aperture), the diffraction is called Fresnel type. device.
(ii) Common examples : Diffraction at a straight (ii) Common examples : Diffraction at single slit,
edge, narrow wire or small opaque disc etc. double slit and diffraction grating.
S Screen Source Screen
at ∞
Source
Slit Slit
(2) Diffraction of light at a single slit : In case of diffraction at a single slit, we get a central bright band
with alternate bright (maxima) and dark (minima) bands of decreasing intensity as shown
P I
θ
x
d S1 θ θ x + ∆x
S2 d sinθ
3λ 2λ λ O λ 2λ 3λ
D Screen − −− ddd
Slit
d dd
(i) Width of central maxima β0 = 2λD ; and angular width = 2λ
d d
(ii) Minima occurs at a point on either side of the central maxima, such that the path difference between the
waves from the two ends of the aperture is given by ∆ = nλ ; where n = 1, 2, 3....
i.e. d sinθ = nλ ⇒ sinθ = nλ
d
(iii) The secondary maxima occurs, where the path difference between the waves from the two ends of the
aperture is given by ∆ = (2n + 1) λ ; where n = 1, 2, 3....
2
i.e. d sinθ = (2n + 1) λ ⇒ sinθ = (2n + 1)λ
2 2d
(3) Comparison between interference and diffraction
Interference Diffraction
Results due to the superposition of waves from two Results due to the superposition of wavelets from
coherrent sources.
different parts of same wave front. (single coherent
source)
All fringes are of same width β = λD All secondary fringes are of same width but the central
d
maximum is of double the width
β0 = 2β = 2 λD
d
All fringes are of same intensity Intensity decreases as the order of maximum increases.
Intensity of all minimum may be zero Intensity of minima is not zero.
Positions of nth maxima and minima Positions of nth secondary maxima and minima
x n(Bright) = nλD , x n (Dark) = (2n − 1) λD x n (Bright) = (2n + 1) λD , x n (Dark) = nλD
d d d d
Path difference for nth maxima ∆ = nλ for nth secondary maxima ∆ = (2n + 1) λ
2
Path difference for nth minima ∆ = (2n − 1)λ Path difference for nth minima ∆ = nλ
(4) Diffraction and optical instruments : The objective lens of optical instrument like telescope or
microscope etc. acts like a circular aperture. Due to diffraction of light at a circular aperture, a converging lens
cannot form a point image of an object rather it produces a brighter disc known as Airy disc surrounded by alternate
dark and bright concentric rings.
The angular half width of Airy disc = θ = 1.22λ (where D = aperture of lens)
D
The lateral width of the image = fθ (where f = focal length of the lens)
Note : ≅Diffraction of light limits the ability of optical instruments to form
clear images of objects when they are close to each other.
(5) Diffraction grating : Consists of large number of equally spaced parallel slits. If light is incident normally
on a transmission grating, the diffraction of principle maxima (PM) is given by d sinθ = nλ ; where d = distance
between two consecutive slits and is called grating element.
Light
θ
II I PM Central I PM II PM
maxima
R2 V2 R1 V1 V1 R1 V2 R2
Polarisation of Light
Light propagates as transverse EM waves. The magnitude of electric field is much larger as compared to
magnitude of magnetic field. We generally prefer to describe light as electric field oscillations.
(1) Unpolarised light
The light having electric field oscillations in all directions in the plane perpendicular to the direction of
propagation is called Unpolarised light. The oscillation may be resolved into horizontal and vertical component.
Direction of
propagation
Direction of propagation Vertical oscillation Horizontal
ll
(2) Polarised light
The light having oscillations only in one plane is called Polarised or plane polarised light.
(i) The plane in which oscillation occurs in the polarised light is called plane of oscillation.
(ii) The plane perpendicular to the plane of oscillation is called plane of polarisation.
(iii) Light can be polarised by transmitting through certain crystals such as tourmaline or polaroids.
(3) Polaroids
It is a device used to produce the plane polarised light. It is based on the principle of selective absorption and
is more effective than the tourmaline crystal. or
It is a thin film of ultramicroscopic crystals of quinine idosulphate with their optic axis parallel to each other.
Polaroid
Plane polarised light
(i) Polaroids allow the light oscillations parallel to the transmission axis pass through them.
(ii) The crystal or polaroid on which unpolarised light is incident is called polariser. Crystal or polaroid on
which polarised light is incident is called analyser.
PA PA
No light
Ordinary light Polarised light Polarised light Ordinary light Polarised light
Transmission axes of the polariser and analyser are parallel to each Transmission axis of the analyser is perpendicular to the
other, so whole of the polarised light passes through analyser polariser, hence no light passes through the analyser
Note : ≅ When unpolarised light is incident on the polariser, the intensity of the transmitted polarised light is
half the intensity of unpolarised light.
(4) Malus law
This law states that the intensity of the polarised light transmitted through the analyser varies as the square of
the cosine of the angle between the plane of transmission of the analyser and the plane of the polariser.
P Intensity = I0 A
Amplitude = A0
Ii Intensity = I
Amplitude = A
θ
(i) I = I0 cos 2 θ and A2 = A02 cos 2 θ ⇒ A = A0 cosθ
If θ = 0o , I = I0 , A = A0 , If θ = 45o , I = I0 , A = A0 , If θ = 90o , I = 0 , A = 0
2 2
(ii) If Ii = Intensity of unpolarised light.
So I0 = Ii i.e. if an unpolarised light is converted into plane polarised light (say by passing it through a
2
plaroid or a Nicol-prism), its intensity becomes half. and I = Ii cos 2 θ
2
Note : ≅Percentage of polarisation = (I max − I min ) × 100
(I max + I min )
(5) Brewster’s law : Brewster discovered that when a beam of unpolarised light is reflected from a
transparent medium (refractive index =µ), the reflected light is completely plane polarised at a certain angle of
incidence (called the angle of polarisation θ p ). Unpolarised θP θP Plane polarised
Also µ = tanθ p Brewster’s law light light
(i) For i < θP or i > θP Partial
Both reflected and refracted rays polarised
becomes partially polarised µ light
(ii) For glass θ P ≈ 57o , for water θ P ≈ 53o Polarisation by reflection
(6) Optical activity and specific rotation
When plane polarised light passes through certain substances, the plane of polarisation of the light is rotated
about the direction of propagation of light through a certain angle. This phenomenon is called optical activity or
optical rotation and the substances optically active.
If the optically active substance rotates the plane of polarisation clockwise (looking against the direction of
light), it is said to be dextro-rotatory or right-handed. However, if the substance rotates the plane of polarisation
anti-clockwise, it is called laevo-rotatory or left-handed.
Polariser Analyser Laevo-rotatory
Substance
Unpolarised light Polarised light dextro-rotatory
Polarimeter
The optical activity of a substance is related to the asymmetry of the molecule or crystal as a whole, e.g., a
solution of cane-sugar is dextro-rotatory due to asymmetrical molecular structure while crystals of quartz are dextro
or laevo-rotatory due to structural asymmetry which vanishes when quartz is fused.
Optical activity of a substance is measured with help of polarimeter in terms of 'specific rotation' which is
defined as the rotation produced by a solution of length 10 cm (1 dm) and of unit concentration (i.e. 1 g/cc) for a
given wavelength of light at a given temperature. i.e. [α ]λt o C = θ where θ is the rotation in length L at
L×C
concentration C.
(7) Applications and uses of polarisation
(i) By determining the polarising angle and using Brewster's law, i.e. µ = tanθP, refractive index of dark
transparent substance can be determined.
(ii) It is used to reduce glare.
(iii) In calculators and watches, numbers and letters are formed by liquid crystals through polarisation of light
called liquid crystal display (LCD).
(iv) In CD player polarised laser beam acts as needle for producing sound from compact disc which is an
encoded digital format.
(v) It has also been used in recording and reproducing three-dimensional pictures.
(vi) Polarisation of scattered sunlight is used for navigation in solar-compass in polar regions.
(vii) Polarised light is used in optical stress analysis known as 'photoelasticity'.
(viii) Polarisation is also used to study asymmetries in molecules and crystals through the phenomenon of
'optical activity'.
Electrons, Photons and X-rays
Electric Discharge Through Gases.
At normal atmospheric pressure, the gases are poor conductor of electricity. If we establish a potential
difference (of the order of 30 kV) between two electrodes placed in air at a distance of few cm from each other,
electric conduction starts in the form of sparks.
The passage of electric current through air is called electric discharge through the air.
The discharge of electricity through gases can be systematically studied with the help of discharge tube shown
below High
– potential +
difference
Length of discharge tube ≈ 30 to 40 cm
Gas Diameter of the tube ≈ 4cm
Manometer
Vacuum pump
The discharge tube is filled with the gas through which discharge is to be studied. The pressure of the enclosed
gas can be reduced with the help of a vacuum pump and it's value is read by manometer.
Sequence of phenomenon
As the pressure inside the discharge tube is gradually reduced, the following is the sequence of phenomenon
that are observed.
– Streamers + – Positive column + Negative glow Positive column
– +
10 mm of Hg Below 4 mm of Hg F.D.S. 1.65 mm of Hg
Cathode glow Negative glow
Negative glow Positive column
– + –+
Cathode glow C.D.S. F.D.S. C.D.S. F.D.S. Striations Greenish light
0.8 mm of Hg 0.05 mm of Hg 0.01 mm of Hg
(1) At normal pressure no discharge takes place.
(2) At the pressure 10 mm of Hg, a zig-zag thin red spark runs from one electrode to other and cracking sound
is heard.
(3) At the pressure 4 mm. of Hg, an illumination is observed at the electrodes and the rest of the tube appears
dark. This type of discharge is called dark discharge.
(4) When the pressure falls below 4 mm of Hg then the whole tube is filled with bright light called positive
column and colour of light depends upon the nature of gas in the tube as shown in the following table.
Gas Colour
Air Purple red
H2
N2 Blue
Cl2 Red
CO2 Green
Na Bluish white
Neon Yellow
Dark red
(5) At a pressure of 1.65 mm of Hg :
(i) Sky colour light is produced at the cathode it is called as negative glow.
(ii) Positive column shrinks towards the anode and the dark space between positive column and negative glow
is called Faradays dark space (FDS)
(6) At a pressure of 0.8 mm Hg : At this pressure, negative glow is detached from the cathode and moves
towards the anode. The dark space created between cathode and negative glow is called as Crook's dark space
length of positive column further reduced. A glow appear at cathode called cathode glow.
(7) At a pressure of 0.05 mm of Hg : The positive column splits into dark and bright disc of light called striations.
(8) At the pressure of 0.01 or 10–2 mm of Hg some invisible particle move from cathode which on striking with
the glass tube of the opposite side of cathode cause the tube to glow. These invisible rays emerging from cathode
are called cathode rays.
(9) Finally when pressure drops to nearly 10–4 mm of Hg, there is no discharge in tube.
Cathode Rays.
Cathode rays, discovered by sir Willium Crooke are the stream of electrons. They can be produced by using a
discharge tube containing gas at a low pressure of the order of 10–2 mm of Hg. At this pressure the gas molecules
ionise and the emitted electrons travel towards positive potential of anode. The positive ions hit the cathode to
cause emission of electrons from cathode. These electrons also move towards anode. Thus the cathode rays in the
discharge tube are the electrons produced due to ionisation of gas and that emitted by cathode due to collision of
positive ions.
(1) Properties of cathode rays
(i) Cathode rays travel in straight lines (cast shadows of objects placed in their path)
(ii) Cathode rays emit normally from the cathode surface. Their direction is independent of the position of the anode.
(iii) Cathode rays exert mechanical force on the objects they strike.
(iv) Cathode rays produce heat when they strikes a material surface.
(v) Cathode rays produce fluorescence.
(vi) When cathode rays strike a solid object, specially a metal of high atomic weight and high melting point X-
rays are emitted from the objects.
(vii) Cathode rays are deflected by an electric field and also by a magnetic field.
(viii) Cathode rays ionise the gases through which they are passed.
(ix) Cathode rays can penetrate through thin foils of metal.
(x) Cathode rays are found to have velocity ranging 1 th to 1 th of velocity of light.
30 10
(2) J.J. Thomson's method to determine specific charge of electron
It's working is based on the fact that if a beam of electron is subjected to the crossed electric field E and
magnetic field B , it experiences a force due to each field. In case the forces on the electrons in the electron beam
due to these fields are equal and opposite, the beam remains undeflected.
S
CA X+ P′
Y– P
L.T.
(H.T.) Magnetic P′′
field S′
C = Cathode, A = Anode, F = Filament, LT = Battery to heat the filament, V = potential difference to
accelerate the electrons, SS' = ZnS coated screen, XY = metallic plates (Electric field produced between them)
(i) When no field is applied, the electron beam produces illuminations at point P.
(ii) In the presence of any field (electric and magnetic) electron beam deflected up or down (illumination at P' or P' ' )
(iii) If both the fields are applied simultaneously and adjusted such that electron beam passes undeflected and
produces illumination at point P.
In this case; Electric force = Magnetic force ⇒ eE = evB ⇒ v = E ; v = velocity of electron
B
As electron beam accelerated from cathode to anode its potential energy at the cathode appears as gain in the
K.E. at the anode. If suppose V is the potential difference between cathode and anode then, potential energy = eV
And gain in kinetic energy at anode will be K.E. = 1 mv 2 i.e. eV = 1 mv 2 ⇒ e v2 ⇒ e E2
2 2 m = 2V m = 2VB 2
Thomson found, e = 1.77 × 1011 C / kg.
m
Note : ≅The deflection of an electron in a purely electric field is given by y= 1 eE . l2 ; where l length of
2 m v2
each plate, y = deflection of electron in the field region, v = speed of the electron.
+ y
→ e–
E –
l
Positive Rays.
Positive rays are sometimes known as the canal rays. These were discovered by Goldstein. If the cathode of a
discharge tube has holes in it and the pressure of the gas is around 10–3 mm of
Hg then faint luminous glow comes out from each hole on the backside of the ←⊕ ⊕→ ←⊕ ⊕→
cathode. It is said positive rays which are coming out from the holes. ←⊕ ←⊕ ←⊕ ⊕→
←⊕ ⊕→ ←⊕
Positive rays
(1) Origin of positive rays
When potential difference is applied across the electrodes, electrons are emitted from the cathode. As they
move towards anode, they gain energy. These energetic electrons when collide with the atoms of the gas in the
discharge tube, they ionize the atoms. The positive ions so formed at various places between cathode and anode,
travel towards the cathode. Since during their motion, the positive ions when reach the cathode, some pass through
the holes in the cathode. These streams are the positive rays.
(2) Properties of positive rays
(i) These are positive ions having same mass if the experimental gas does not have isotopes. However if the
gas has isotopes then positive rays are group of positive ions having different masses.
(ii) They travels in straight lines and cast shadows of objects placed in their path. But the speed of the positive
rays is much smaller than that of cathode rays.
(iii) They are deflected by electric and magnetic fields but the deflections are small as compared to that for
cathode rays.
(iv) They show a spectrum of velocities. Different positive ions move with different velocities. Being heavy,
their velocity is much less than that of cathode rays.
(v) q /m ratio of these rays depends on the nature of the gas in the tube (while in case of the cathode rays q/m
is constant and doesn't depend on the gas in the tube). q/m for hydrogen is maximum.
(vi) They carry energy and momentum. The kinetic energy of positive rays is more than that of cathode rays.
(vii) The value of charge on positive rays is an integral multiple of electronic charge.
(viii) They cause ionisation (which is much more than that produced by cathode rays).
Mass Spectrograph.
It is a device used to determine the mass or (q/m) of positive ions.
(1) Thomson mass spectrograph
It is used to measure atomic masses of various isotopes in gas. This is done by measuring q/m of singly ionised
positive ions of the gas.
Low Cathode – S – Screen or S Y
pressure P + Photo plate y
gas +q v z
Q Z
N N
D
Pump
The positive ions are produced in the bulb at the left hand side. These ions are accelerated towards cathode.
Some of the positive ions pass through the fine hole in the cathode. This sfcinreeenra(yEo|f|pBosbituivteEioonrs Bis subjected to
electric field E and magnetic field B and then allowed to strike a fluorescent ⊥ v ).
If the initial motion of the ions is in + x direction and electric and magnetic fields are applied along + y axis
then force due to electric field is in the direction of y-axis and due to magnetic field it is along z-direction.
The deflection due to electric field alone y = qELD .........(i)
mv 2
The deflection due to magnetic field alone z = qBLD .........(ii)
mv
From equation (i) and (ii)
z2 = k q y , where k = B 2 LD ; This is the equation of parabola. It means all the charged particles moving
m E
with different velocities but of same q/m value will strike the screen placed in yz plane on a parabolic track as shown
in the above figure.
Note : ≅All the positive ions of same. q/m moving with different velocity lie on the same parabola. Higher is
the velocity lower is the value of y and z. The ions of different specific charge will lie on different
parabola. Y q/m q/m q/m q/m
V1 V2 V3 V4 V1>V2>V3>V4 large small small large
Z Light mass
Heavy mass
≅ The number of parabola tells the number of isotopes present in the given ionic beam.
(2) Bainbridge mass spectrograph
In Bainbridge mass spectrograph, field particles of same velocity are selected by using a velocity selector and
then they are subjected to a uniform magnetic field perpendicular to the velocity of the particles. The particles
corresponding to different isotopes follow different circular paths as shown in the figure.
(i) Velocity selector : The positive ions having a certain velocity v gets isolated from all other velocity
particles. In this chamber the electric and magnetic fields are so balanced that the particle moves undeflected. For
this the necessary condition is v = E .
B
(ii) Analysing chamber : In this chamber magnetic field B is applied perpendicular to the direction of motion
of the particle. As a result the particles move along a circular path of Velocity spectrum ••
radius
mv • • ••
• •• →
•
• •B′
mE q E r1 m1 +q • • •
r = qBB' ⇒ m = BB' r also r2 = m2 2r2 2r1 • •• •
r • •
m1 • •
→→ •
In this way the particles of different masses gets deflected on circles •m2 ••
of different radii and reach on different points on the photo plate. E ⊥B
• ••
•
Photographic plate ••
Note : ≅ Separation between two traces = d = 2r2 − 2r1 ⇒ d = 2v(m2 − m1 ) .
qB'
Matter waves (de-Broglie Waves).
According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.
or
A wave is associated with moving material particle which control the particle in every respect.
The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the
form of wave packets with group velocity.
(1) de-Broglie wavelength
According to de-Broglie theory, the wavelength of de-Broglie wave is given by
λ = h = h = h ⇒ λ ∝ 1 ∝ 1 ∝ 1
p mv 2mE p v E
Where h = Plank's constant, m = Mass of the particle, v = Speed of the particle, E = Energy of the particle.
The smallest wavelength whose measurement is possible is that of γ -rays.
The wavelength of matter waves associated with the microscopic particles like electron, proton, neutron,
α-particle etc. is of the order of 10−10 m.
(i) de-Broglie wavelength associated with the charged particles.
The energy of a charged particle accelerated through potential difference V is E = 1 mv 2 = qV
2
Hence de-Broglie wavelength λ = h = h h
p = 2mqV
2mE
λelectron = 12.27 Å, λ proton = 0.286 Å, λdeutron = 0.202 × 10 −10 Å, λα − particle = 0.101 Å
V V V V
(ii) de-Broglie wavelength associated with uncharged particles.
For Neutron de-Broglie wavelength is given as λ Neutron = 0.286 × 10−10 m = 0.286 Å
E (in eV) E (in eV )
Energy of thermal neutrons at ordinary temperature
E = kT ⇒ λ = h ; where k = Boltzman's constant = 1.38 × 10−23 Joules/kelvin , T = Absolute temp.
2mkT
So λThermal Neutron = 6.62 × 10−34 = 30.83 Å
(2) Some graphs
2 × 1.07 × 10−17 × 1.38 × 10−23 T T
λ λλ
Slope = h Small m
p Large m
1/ v
λ Small λ λ Small
Large Large
Small m
1/V Large m 1E
E
Note : ≅ A photon is not a material particle. It is a quanta of energy.
≅When a particle exhibits wave nature, it is associated with a wave packet, rather then a wave.
(3) Characteristics of matter waves
(i) Matter wave represents the probability of finding a particle in space.
(ii) Matter waves are not electromagnetic in nature.
(iii) de-Brogile or matter wave is independent of the charge on the material particle. It means, matter wave of
de-Broglie wave is associated with every moving particle (whether charged or uncharged).
(iv) Practical observation of matter waves is possible only when the de-Broglie wavelength is of the order of
the size of the particles is nature.
(v) Electron microscope works on the basis of de-Broglie waves.
(vi) The electric charge has no effect on the matter waves or their wavelength.
(vii) The phase velocity of the matter waves can be greater than the speed of the light.
(viii) Matter waves can propagate in vacuum, hence they are not mechanical waves.
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