Pra-U STPM Physics Penggal 3 2018 CB039248c

Physics Term 3 STPM Chapter 22 Geometrical Optics 145 22 Example 17 Object O 40.0 cm 20.0 cm 1 2 3 20.0 cm Three lenses and an illuminated object are arranged coaxially as shown in the above fi gure. The two convex lenses have the same focal length 20.0 cm and the focal length of the concave lens is 10. 0 cm. Determine the position and the characteristics of the image formed by (a) lens 1 only, (b) lens 1 and lens 2, (c) all the three lenses. Draw a ray diagram for your answer to (c). Solution: (a) f = +20 cm, u = 40.0 cm = 2f When u = 2f, v = u = 2f = 40.0 cm The image is real and 40.0 cm from lens 1 and on the right side of lens 1. (b) The image I1 , acts as a virtual object for lens 2. u = –20.0 cm f = –10 cm (concave lens) Using 1 u + 1 v = 1 f 1 v = 1 –10 – 1 –20 v = –20.0 cm The image I2 is virtual and 20.0 cm to the left of lens 2. (c) O I1 I2 I3 40.0 cm 1 2 3 The image I 2 acts as a real object for lens 3 u = +40.0 cm f = +20.0 cm Since u = 2f, then v = u = 2f = 40.0 cm The fi nal image I3 is real and 40.0 cm to the right of lens 3. O I1 I2 40.0 cm 1 2 20.0 cm

Physics Term 3 STPM Chapter 22 Geometrical Optics 146 22 Example 18 Two lenses and a convex spherical mirror each of focal length 20.0 cm are arranged coaxially as shown in the fi gure. An object O is 30.0 cm from lens 1. O 30.0 cm 50.0 cm 20.0 cm Lens 1 Object Lens 2 Convex mirror Determine the position and characteristics of the image formed by (a) lens 1 only, (b) lens 1 and lens 2, (c) all the components after light from the object passes through the two lens, refl ected by the convex mirror and passes through the lens again. Illustrate your answer to (c) by a ray diagram. Solution: (a) Focal length of lens 1, f = +20.0 cm Object distance, u = +30.0 cm Using 1 u + 1 v = 1 f 1 v = 1 20 – 1 30 v = 60.0 cm Image is real, twice the height of the object (m = v u = 2.0) and 60.0 cm from lens 1. (b) The image I1 , formed by the lens 1 acts as a virtual object for lens 2. Then u = –10.0 cm f = –20.0 cm (concave lens) using 1 u + 1 v = 1 f 1 v = 1 –20 – 1 –10 v = 20.0 cm The image is real and 20.0 cm to the right of lens 2. (c) O P I1 I2 50.0 cm 10.0 cm 10.0 cm Lens 1 Lens 2 Convex mirror O I1 50.0 cm 10.0 cm 60.0 cm Lens 1 Lens 2

Physics Term 3 STPM Chapter 22 Geometrical Optics 147 22 Important Formulae 1. For spherical mirrors: focal length, f = r 2 , r = radius of curvature 2. Spherical mirror equation and thin lens equation: 1 f = 1 u + 1 v Linear magnifi cation, m = hi h0 = v u = v f – 1 3. Refractive index of medium 1 to medium 2, n1 = sin i sin r n1 sin θ1 = n2 sin θ2 4. Refraction at a spherical surface: n1 u + n2 v = |n2 – n1| r 5. Lens maker equation: 1 f = (n – 1) ( 1 r1 + 1 r2 ) and 1 f = ( n2 n1 – 1) ( 1 r1 + 1 r2 ) STPM PRACTICE 22 1. The radius of curvature of a concave mirror is 10.0 cm. An object is placed 20.0 cm from the mirror. What is the linear magnifi cation of the image? A 1 5 C 1 3 B 1 4 D 1 2 2. An object is 60 cm from a lens. An image of the same size as the object is formed on a screen behind the lens. Which is the lens? A diverging lens of focal length 30 cm B diverging lens of focal length 15 cm C converging lens of focal length 15 cm D converging lens of focal length 30 cm The image I2 , is at the pole P of the convex mirror. It acts as an object for the mirror, with u = 0. Hence the image which is a spot of light is at P. The spot of light at P now acts as a real object for the concave lens. Light refl ected by the mirror passes through the lens with u = +20.0 cm f = –20.0 cm and 1 v = 1 –20 – 1 –20 v = –10.0 cm which is at I1 (from (b) above) The refl ected light after passing through lens 2 seems to come from I1 , which acts as a real object for lens 1 with u = +60.0 cm f = +20.0 cm 1 v = 1 20 – 1 60 v = +30.0 cm The refl ected light after passing lens 2 is focussed by lens 1 at O, the position of the object. Hence, the fi nal image is at the same position as the object.

Physics Term 3 STPM Chapter 22 Geometrical Optics 148 22 3. For which position of the object is the image formed by a concave mirror virtual? A Any position B Object distance less than the focal length C Object distance less than twice the focal length D Object distance greater than twice the focal length 4. A plano-convex lens of refractive index 1.50 is placed in a dish of water as shown in the figure. Glass Water The curve surface of the lens has a radius of curvature of 20 cm. What is the focal length of the combination of glass and water lenses? [Refractive index of water = 4 3 ] A –20 cm C 60 cm B 20 cm D 120 cm 5. A thin lens made from a material of refractive index 1.48 is shown in the diagram below. 18.0 cm Object It has spherical surfaces of radii 5.00 cm and 3.50 cm. What is the nature and position of an image formed when an object is placed 18.0 cm to the left of the lens? A A real image at 5.63 cm to the right of the lens. B A real image at 10.3 cm to the right of the lens. C A virtual image at 5.63 cm to the left of the lens. D A virtual image at 10.3 cm to the left of the lens. 6. A concave mirror produces a virtual image which is three times the height of the object at a distance of 60 cm from the mirror. What is the focal length of the concave mirror? A 10 cm C 30 cm B 20 cm D 40 cm 7. v n O r 2r u I The figure shows a glass rod of diameter 2r. One end of the rod has a surface which is a hemisphere of radius r. The refractive index of the glass is n. A point object O is at a distance u from the end of the rod. The image I due to refraction is at a distance of v, which is twice the object distance u from the end of the rod. Which of the following is equal to u? A (n + 2)r n C 2(n – 1)r (n + 2) B (n + 2)r 2(n – 1) D 2(n – 1)r (n – 2) 8. Parallel rays are incident on a spherical water droplet of radius 4 mm. After refraction by the water droplet, the rays are brought to a focus a distance f from the centre of the sphere. If the refraction index of water is 4 3 , the value of f is A 8 mm C 5 mm B 4 mm D 6 mm 9. The figure shows a convex lens of focal length 15.0 cm positioned on the surface of water in container. Parallel rays are incident on the lens surface. Parallel rays Lens Air Water If the refractive index of the material of the lens is greater than that of water, the rays that emerge from the lens A are brought to a focus at a distance of 15 cm from the lens B are brought to a focus at a distance less than 15 cm from the lens C are brought to a focus at a distance more than 15 cm from the lens D diverge from a point above the lens

Physics Term 3 STPM Chapter 22 Geometrical Optics 149 22 10. The minimum distance between a real object and its real image formed by a convex lens is d. What is the distance of the object from the lens? A d 4 C d 2 B d 3 D 3 4 d 11. A luminous object is at a distance of 90 cm from a screen. To obtain a sharp image twice the size of the object on the screen, the lens placed between the object and screen is a lens of focal length A f = – 60 cm C f = +30 cm B f = +20 cm D f = –10 cm 12. O P Q R T S I A point object O is moved from the principal axis of a convex lens along the direction OP as shown in the above figure. The image I would move in the direction A IQ C IS B IR D IT 13. A beam of light is directed towards a convex lens of focal length 20 cm. The beam is parallel to the principal axis of the lens. A rectangular aperture of side 1.0 cm is placed in front of the lens. A screen is placed 30 cm behind the lens as shown in the figure. f = 20 cm 30 cm Aperture of sides 1.0 cm Opaque card Screen Which of the following shows correctly the triangular spot of light on the screen? A 1.5 cm C 0.5 cm B 1.0 cm D 0.5 cm 14. A convex lens and a concave lens which are placed coaxilly at a distance of 20.0 cm apart is shown in the diagram below. Convex lens Concave lens 20.0 cm Each lens has a focal length of 30.0 cm. If patrallel rays of light from a distant object are incident on the convex lens, what is the position of the final image formed after the rays pass through the two lenses? A 7.5 cm left of the concave lens B 7.5 cm right of the concave lens C 15.0 cm left of the concave lens D 15.0 cm right of the concave lens 15. Which graph shows the variation of the reciprocal of magnification 1 m of an object with distance u of the object from a convex lens? A 0 m u − 1 C 0 m u − 1 B 0 m u − 1 D 0 m u − 1 16. A concave mirror of radius of curvature r produces an upright image which is twice the height of an object placed in front of the mirror. What is the distance of the object from the mirror? A 1 4 r C 3 4 r B 1 2 r D r

Physics Term 3 STPM Chapter 22 Geometrical Optics 150 22 17. An object O, a converging lens L1 of focal length 10.0 cm, and a meniscus lens L2 are arranged as shown in the diagram. The radii of curvature of the meniscus lens are 5.0 cm and 7.5 cm. 20.0 cm 35.0 cm L1 L2 O (i) Find the focal length of the meniscus lens. (ii) Where is the final image, and its magnification? (iii) Draw a ray diagram to show the formation of the final image. 18. (a) The surfaces of a biconvex glass lens have the same radius of curvature, 20.0 cm. The refractive index of glass is 1.50. An object is placed 30.0 cm from the lens. Find (i) the focal length of the lens, (ii) the position of the image, (iii) the linear magnification. State the characteristics of the image. (b) A plane mirror is then placed behind the lens with its reflecting surface in contact with the lens, (i) Determine the position of the image formed by the lens and mirror. (ii) What is the overall magnification? (iii) State the characteristics of the final image. (iv) Draw a ray diagram to show the image formed by the lens, and the final image. 19. (a) A concave mirror has a radius of curvature of 16.00 cm. An object of height 2.50 cm is placed at 20.00 cm in front of the mirror. (i) Determine the position and size of the image, and state its nature. (ii) Sketch a ray diagram to show the formation of the image. (b) An observer who is looking at a fish inside a thin spherical glass bowl of water is shown in diagram below. Water 18.0 cm 16.0 cm Spherical glass bowl The thin spherical glass bowl has a diameter of 32.0 cm and the refractive index of water is 1.33. If the fish is at 18.0 cm from the near wall on the diameter of the bowl, determine the location of the fish as seen by the observer. Neglect the glass wall of the bowl. (c) The concave mirror in (a) is attached to the outside wall of the far side of the fish bowl with respect to the observer. Determine the location of the final image of the fish as seen by the observer. 20. An object AB is in front of a concave mirror as shown in the figure. The point F is the focal point of the mirror. A B F (a) Draw a ray diagram to locate the position of the image of the point A of the object. Then draw the image of the object AB formed by the mirror. (b) State the characteristics of the image. 21. (a) The power of a lens made of glass of refractive index 1.50 is +4.00 D. One of the lens surfaces is convex and of radius of curvature 20.0 cm. (i) What is the focal length of the lens? (ii) Determine the radius of curvature of the other surface of the lens. Is this surface convex or concave?

Physics Term 3 STPM Chapter 22 Geometrical Optics 151 22 (iii) Discuss the change in the power of the lens when it is place in a liquid of refractive index higher than the glass of the lens. (b) A microscope has an objective lens of focal length 1.60 cm and an eyepiece of focal length 6.50 cm. When a sample is placed 3.20 cm from the objective lens, a final image is formed when the separate between the lenses is 8.35 cm. 8.65 cm Objective lens Eyepiece Determine (i) the position of the final image (ii) the magnification of the image 22. (a) A concave mirror of focal length 10.0 cm is placed in a beaker containing a liquid up to a depth of 7.5 cm. The refractive index of the liquid is 1.2. Find the distance of an optical pin above the liquid surface when there is no parallax between the pin and its image. (b) (i) A pin P of length 3.0 cm is placed in a beaker. A converging lens L of focal length 10.0 cm is held 15.0 cm above the pin. Find the position and size of the image. (ii) Water of refracting index 4 3 is poured into the beaker to a depth of 10.0 cm. Find the new position of the image. Discuss any change to the size of the image. 23. (a) A plano-concave lens has a surface of radius of curvature 50.0 cm. If the refractive index of the material of the lens is 1.50, calculate the focal length of the lens (i) in air, (ii) in water of refractive index 1.33. (b) A camera uses a single convex lens of focal length 50 mm. The lens is able to produce images of objects from a distance of 1.0 m to infinity from the lens. Calculate the distance the lens need to be moved to take pictures of objects between the distances mentioned. 24. (a) Two convex lenses have focal lengths 10.0 cm and 30.0 cm respectively. When a concave lens is arranged in contact with one of the convex lenses the focal length of the combination is 17.5 cm. (i) Calculate the focal length of the concave lens. (ii) The concave lens is then arranged in contact with the other convex lens. What is the focal length of the combination. Explain whether the lens combination is converging or diverging? (b) Two thin lenses of focal length +10.0 cm and –10.0 cm are arranged coaxially and separated by a distance of 7.0 cm. An object is placed on the axis at a distance of 43.5 cm from the centre of the system of lenses, first on the left side, and then on the right side. Find the position of the final image for each position of the object. 25. (a) Write an expression relating the object distance, u the image distance v and the focal length f of a concave lens. Explain the sign convention you use to determined whether a distance is positive or negative. (b) An illuminated object, a convex lens and a convex spherical mirror are arranged coaxially so that light from the object passes through the lens, reflected by the mirror and then passes through the lens again. The lens and curve mirror both has focal length 20.0 cm and the lens is 40.0 cm from the object. D e t e r m i n e t h e p o s i t i o n a n d characteristic of the final image when the curve mirror is (i) in contact with the lens, (ii) 40.0 cm from the lens. Illustrate your answers with ray diagrams. 26. (a) What is meant by the radius of curvature of a convex spherical mirror? (b) The figure shows a spherical soap bubble formed at the end T of a glass tube when air is blown into the tube.

Physics Term 3 STPM Chapter 22 Geometrical Optics 152 1 1. v = +5.0 cm Image is enlarged, upright and virtual. 2. v = –13.3 cm Image is virtual, upright and magnifi ed. 3. r = 30.0 cm m = v u , r = 2f 1 v + 1 u = 1 f 4. (a) v = 21.0 cm Image real, inverted and diminished. (b) v = 30.0 cm Image real, inverted and same size as object. (c) v = –30.0 cm Image virtual, upright and magnifi ed. 5. (a) v = –5.83 cm (b) m = –0.42 6. (a) Concave mirror because only concave mirror can produced a magnifi ed image. Image is virtual because it is upright. (b) r = 40.0 cm 1 v + 1 u = 1 f , r = 2f 7. (a) Image is real, hence concave mirror. (b) r = +13.3 cm. v = 10.0 cm, u = 20.0 cm 8. (a) When there is non-parallax between the optical pin and its image, they are at the same position. There is no relative movement between the optical pin and its image when viewed from diff erent angles. (b) (i) u = v = 40.0 cm 1 v + 1 u = 1 f , 1 f = 1 40.0 + 1 40.0 The radius of the bubble is r. An object O is at a distance L from the end T of the tube. Light is refl ected from the surface of the soap bubble. Glass tube Soap bubble L T O r (i) If the distance of the image of O formed by refl ection of light is at a distance s from the end T, derive an expression for s in terms of L and r. (ii) Hence, calculate the value of s when r = 1 2 L. Explain qualitatively your answer. (c) The figure below shows spherical surface between two media of refractive indices n1 and n2 . The point O is the centre of curvature of the spherical surface and OP = r the radius of curvature. A point object X is at a distance u from P. The image produced by the refraction of light at the spherical surface is at a distance v from P. n2 O P X n1 (i) Write an expression to relate u, v, r, n1 and n2 . Explain clearly the sign convention you use for r, u, and v. (ii) If r = 20.0 cm, u = 15.0 cm, n1 = 1.0 and n2 = 1.5, calculate the value of v and state whether the image is real or virtual. (iii) An image of X due to refl ection of light from the spherical surface is also formed. Calculate the distance between this image and the image in (c) (ii) above. 22 ANSWERS

Physics Term 3 STPM Chapter 22 Geometrical Optics 153 22 Focal length, f = 20.0 cm (ii) Radius of curvature, r = 2f = 40.0 cm (c) 40.0 cm 9. (a) Convex lens Convex mirror Object 15.0 cm 10.0 cm 30.0 cm P Screen (b) The rays after passing through the convex lens must be incident at right-angles on the convex mirror. Then the reflected rays would travel back along the same path to form the final image at the position of the object. Hence distance of the convex mirror from P = r, radius of curvature of the mirror. r = (30.0 – 10.0 cm) = 2f Focal length of convex mirror f = 10.0 cm 2 1. (a) v = ∞ n1 u + n2 v = n2 ~ n1 r , r positive (b) v = 300 cm r positive (c) v = –42.9 cm r negative 2. (a) O A I1 v u = 20.0 cm v = 30.0 cm (b) O B I1 I2 u = –20.0 cm v = BI = 5.71 cm 3. A B I1 I (a) AI1 = 15 cm BI = 2.5 cm, u = –5.0 cm 4. (a) u = 8.0 cm v = –7.2 cm A I O (b) u = 16.0 cm I O B v = BI = –18.0 cm 5. O I r = 9.00 cm n2 = 1.0, n1 = 1.5 u = 3.00 cm v = –2.25 cm 3 1. A: v = u = 2f = 20 cm 2. C: Minimum distance between object and real image = 4f 3. A: Diverging lens does not form real image. Converging lens: when v = u = 2f, minimum distance of u + v =4f 4. D: 4f = 4(0.3) m = 1.2 m. For real image, minimum distance of OI, is 4f. 5. B: For converging lens, positions of real image and object are interchangeable. 6. D: Use 1 f = 1 f 1 + 1 f 2 7. C: OI = 100 cm = 4 (25 cm) = 4f. Hence v = u = 2f, u + v = 4f. 8. B: 5 cm 15 cm 10 cm 20 cm 10 cm A: B: C: D: 9. A: Lens P: 1 v = 1 10 – 1 15, v = 30 cm Lens Q: u’ = (40 – 30) cm = 10 cm = 2f v’ = u’ = 2 f = 10 cm

Physics Term 3 STPM Chapter 22 Geometrical Optics 154 22 10. B: Initially, convex lens: image at focus, v = 10 cm Concave lens: u = –(10 – 5) cm = -5 cm 1 v = 1 –5 – 1 –5, v = ∞ Emerging rays are parallel. When concave lens is nearer to convex lens, u’ = –6 cm (say), then v is negative (rays diverge) 11. D : Use 1 f = 1 u + 1 v gives v1 = –5.0 cm u2 = 10.0 cm gives v2 = 10.0 cm 12. B : 1 f 1 > 1 f 2 , 1 F = 1 f 1 + 1 –f 2 F is positive 13. A : u = f + x1 , v = f + x2 1 f = 1 u + 1 v gives f = x1 x2 14. A: 1 50.0 + 1 (–25.0) = 1 f 15. D 16. (a) (i) O F I Converging lens (ii) v = 7.5 cm, m = 0.5 Image is real and inverted (b) (i) f = r 2 = – 20.0 2 cm 1 –10.0 = 1 15.0 + 1 v v = – 6.0 cm (ii) hI = ( 6.0 15.0)(5.0) cm = 2.0 cm (c) (i) Converging lens. (ii) v = 1 2 u, u= 2v u + v = 60.0 cm v = 20.0 cm 17. (a) (i) –120.0 cm r1 = –20.0 cm, r2 = +30.0 cm (ii) –469 cm (b) Power or focal length of lens changes. 18. v = –16.0 cm Image virtual, upright and diminished 19. 43.3 cm above the lens. 20. O 6 cm 50.0 cm 20.0 cm 30.0 cm I I1 Image 6.0 cm from concave lens (a) v1 = 40.0 cm (b) Final image at infinity STPM Practice 22 1. C: f = 1 2 r = 5.0 cm, 1 v = 1 f – 1 u = 1 5.0 – 1 20.0 v = 3 20 m = v u = 20/3 20 = 1 3 2. D: When real image is the same size as the object, u = v = 2f = 60 cm f = + 30 cm 3. B 4. D: Glass lens, 1 f 1 = (1.50 – 1)( 1 ∞ + 1 20), f1 = +40 cm Water lens, 1 f 2 = ( 4 3 – 1)( 1 ∞ – 1 20), f 1 = –60 cm 1 f = 1 40 + 1 –60 , f = 120 cm 5. D: 1 f = (n – 1)( 1 r1 + 1 r2 ) = (1.48 – 1)( 1 +5.0 + 1 –3.50 ) = –0.0411 1 v = 1 f – 1 u = (–0.0411) – 1 18.0 v = –10.3 cm 6. C: v = - 3u = – 60 cm, u = 20 cm 1 f = 1 u – 1 v = 1 20 – 1 –60, f = 30 cm 7. B: Use n1 u + n2 v = n2 – n1 r 1 u + n 2u = n – 1 r 2 + n 2u = n – 1 r u = (n + 2)r 2(n – 1) 8. D: n1 u + n2 v = n2 – n1 r First refracting surface, 1 ∞ + 4/3 v1 = 4/3 – 1 4 , v1 = 16 mm I I 1 v v1

Physics Term 3 STPM Chapter 22 Geometrical Optics 155 22 Second refracting surface, u = –(16 – 8) = –8 mm 4/3 –8 + 1 v = 4/3 – 1 4 , v = 4 mm = (4 + 4) mm from the centre (f = 8 mm) 9. C: Focal length of lens < 15 cm 10. C: Minimum object-image distance = 4f when u = v = 2f = d 2 11. B: v = 2u, v + u = 90 cm, hence u = 30 cm, v = 60 cm 1 f = 1 30 + 1 60 , f =+ 20 cm 12. C: Image is inverted. When u decreases, v increases. 13. D: Image is inverted. 20 cm ƒ = 20 cm 10 cm 14. D: Image I1 : 30.0 cm from convex lens, and acts as virtual object. For concave lens, u = –10.0 cm 1 –30.0 = 1 –10.0 + 1 v2 , gives v2 = +15.0 cm to the right of concave lens. 15. A: 1 f = 1 u + 1 v , 1 m = u v = u f – 1. When u = f, 1 m = 0 16. A : f = r 2 , v = –2u 17. (i) r1 = 5.0 cm r 2 = –7.50 cm Use 1 f = (n – 1)( 1 r1 + 1 r2 ) f = 30.0 cm (ii) Lens L1 , u = 20.0 cm = 2f 1 gives v = u = 20.0 cm Lens L2 , u2 = (35.0 – 20.0) cm gives v2 = –30.0 cm (30.0 cm in front of L2 ) Magnification = ( v u ) ( v2 u ) = - 2.00 (iii) Hint: Draw ray diagram on graph paper. L1 L2 I 1 O 18. (a) (i) 1 f = (n – 1)( 1 r1 + 1 r2 ) = (1.50 – 1)( 1 20 + 1 20) f = 20.0 cm (ii) 1 v = 1 f – 1 u = 1 20.0 – 1 30.0, v = 60.0 cm (iii) m = v u = 60.0 30.0 = 2.00 Image is real, inverted and magnified. (b) (i) With the mirror behind the lens, the first image acts as a virtual object of the Lens, u = - 60.0 cm 1 v = 1 f – 1 u = 1 20.0 – 1 –60.0, v = + 15.0 cm (ii) Overall magnification m1 m2 = (2.00) ( 15.0 60.0 ) = 0.50 (iii) Final image is real and diminished. (iv) O I 2 I 1 19. (a) (i) f = 1 2 r = 8.00 cm, u = 20.0 cm 1 u + 1 v = 1 f , 1 v = 1 8.00 – 1 20.0 v = 13.3 cm m = hi ho = v u, Height of image, hi = ( 13.3 20.0)(2.50) cm = 1.66 cm Image is real, inverted and smaller than the object. (ii) F O I (b) (i) n1 u + n2 v = n2 – n1 r , 1.33 18.0 + 1.00 v = 1.33 – 1.00 16.0 v = –18.8 cm (ii) Concave mirror, f = 8.0 cm, u = (32.00 – 18.8) cm = 13.2 cm 1 u + 1 v = 1 f , 1 v = 1 8.0 – 1 13.2 v = 20.3 cm

Physics Term 3 STPM Chapter 22 Geometrical Optics 156 22 This image acts as a real object for the refraction at the left surface of the bowl. 1.33 (32.0 – 20.3) + 1.00 v = 1.33 –1.00 16.0 , v = –10.7 cm Final image is 10.7 cm to the right of the left refracting surface of the bowl. Mirror 20.3 cm 13.2 cm 18.8 cm 11.7 cm 10.7 cm I I 2 I O 1 O: Object I1 : Image in (i) I2 : Image formed by mirror I : Final image seen by observer 20. (a) Mirror A B F IA IB (b) Image is real, inverted and magnified (twice). 21. (a) Power 1 f , f = 1 4.00m = 0.250 cm 1 f = 0.250 = (1.50 – 1)( 1 0.200 – 1 r2 ) r2 = 0.222 m (convex) (b) (i) f 0 = 1.60 cm u1 = 3.20 cm = 2f 0 Hence v1 = 2f 0 = 3.20 cm u2 = (8.35 – 3.20 )cm = 5.15 cm 1 v2 = 1 6.50 – 1 5.15 v2 = – 24.8 cm Image is virtual and in front of eyepiece. (ii) m = m1 m2 = (3.20 3.20)(–24.8 5.15 ) = –4.82 22. (a) 10.4 cm (b) (i) 30.0 cm, 6.0 m (ii) 50.0 cm Image is enlarged 23. (a) (i) f = –100.0 cm r1 = ∞, r2 = –50 cm (ii) f = –391 cm (b) 2.63 mm u = ∞, v = 50 mm u = 1 m, v = 52.6 mm 24. (a) f = –23.3 cm 1 F = 1 f 1 + 1 f 2 f = –104 cm diverging (b) Object in front of convex lens, image 20.5 cm from centre of lens system. Object in front of concave lens, image 33.5 cm from centre of lens system. 25. (i) Image is real and at the same position as the object. O 40 cm I I 1 40 cm (ii) Image is real and at the same position as the object. O 40 cm I 40 cm 26. (a) Refer to page 116 (b) (i) s = r(3L – 4r) 2L – 3r f = – r 2 (convex mirror) v = – (2r – s) u = (L – 2r) 1 f = 1 u + 1 v O I L s T (ii) r = 1 2 L ⇒ s = L Object O on surface of bubble, u = 0 (c) (i) n1 u + n2 v = n2 ~ n1 r (ii) –36.0 cm Image virtual (iii) v′ = –6.0 cm, distance II′ = 42.0 cm O I′ I 6.0 cm (Reflection) (Refraction) 36.0 cm

157 CHAPTER WAVE OPTICS 23 Concept Map 157 Interference Huygens’s Principle Optical Waveguides Fibre optics and Waveguards Two-slit Interference x = λD a Diffraction by Single Slit sin θ = λ a Polarisation Using Polaroid Malus’s law I = I0 cos2 θ By Refl ection θΒ Brewster’s law tan θB = n Interference in Thin Film t = λ 4µ Monochromatic light Thin film O A B C X Y t Diffraction Gratings d sin θ = mλ Wave Optics Bilingual Keywords Constructive interference: Interferens membina Destructive interference: Interferens memusnah Diffraction grating: Parutan belauan Diffraction: Belauan Order of diffraction: Peringkat belauan Path difference: Beza lintasan Polarisation: Pengutuban Slit: Celah

Physics Term 3 STPM Chapter 23 Wave Optics 158 23 23.1 Huygens’s Principle Learning Outcomes Students should be able to: • state the Huygens’s principle • use the Huygens’s principle to explain interference and diffraction phenomena 1. Huygens’s wave theory of light is based on geometrical construction that enables us to determine the position of a wavefront at any time t after the present position. 2. A wavefront is a surface where all points of the surface oscillate in phase. 3. In a homogenous medium, the direction of propagation of a wave is perpendicular to the wavefront. Plane wavefront (a) (b) Direction of propagation Spherical wavefront S Direction of propagation (a) (b) Figure 23.1 4. Figure 23.1(a) shows plane wavefronts, and Figure 23.1(b) shows a section of a spherical wavefronts from a point source S. 5. Huygens’s principle states that all points on a wavefront serve as point sources of spherical secondary wavelets. After a time t, the new position of the wavefront is the surface that is tangential to these secondary wavelets. INTRODUCTION 1. Although light shows properties such as refl ected and refraction, the idea that light propagates in the form of waves as suggested by Huygens was initially not accepted by most scientists. 2. This is because the scientists were unable to demonstrate that light exhibits wave properties such as interference and diffraction. 3. Newton himself suggested in his corpuscular theory of light that light propagates in the form of particles which he called corpuscules. 4. In 1801, Thomas Young showed the interference of light using a double-slit. Only then Huygens’s wave principle of light was accepted.

Physics Term 3 STPM Chapter 23 Wave Optics 159 23 P R Q ct ct S a b c d e Direction of propagation Time = 0 Time = t Figure 23.2 6. Figure 23.2 shows a plane wavefront PQ at time t = 0. To determine the position of the wavefront after a time = t using Huygens’s principle, mark points like a, b, c, d, e, ... on the wavefront. Draw arcs which represent the secondary wavelets which are in phase with each other from a, b, c, d, e, ... After a time = t, wavelets from these points a, b, c, ... would have travel a distance ct, where c is the speed of light. The plane RS that is tangent to all these wavelets represents the wavefront at time = t. Example 1 A fl ash of light is produced at a point on the surface of the Earth at a distance 1.5 m in front of a large vertical mirror. The fl ash lasts for 10–9 s. Use Huygens’s principle to draw a diagram to show the light wavefront (a) 5 × 10–9 s (b) 1 × 10–8 s after the fl ash. Show clearly the centres of any arcs that you draw. Solution: 7 6 (a) (b) (b) 1 2 3 4 5 O 1.5 m Source I

Physics Term 3 STPM Chapter 23 Wave Optics 160 23 Explanation (a) Distance travelled by light in 5 × 10–9 s = 3.0 × 108 × (5 × 10–9) = 1.5 m Hence the light wavefront is a semicircle of 1.5 m centre at O, the source (b) To obtain the position of the light wavefront after a further time of 5 × 10–9 s or 1 × 10–8 s from the start, draw arcs of radius 1.5 m (to scale) centre at points marked 1, 2, 3, 4, 5 and 6. The arc of radius 3.0 m (to scale) centre at O is tangent to the wavelets from point 1, 2, 3, 4, and 5. The arc of radius 3.0 m (to scale) centre at the point I is tangent to wavelets from points such as 6 and 7. This represent the wavefront refl ected by the mirror. Quick Check 1 1. Direction of propagation Figure (a) (a) In Figure (a), the three parallel lines represent the wavefronts of a parallel beam of light at time t = 0, t = T, and t = 2T. Use Huygens’s principle to construct the wavefront at time t = 3T. A B Direction of propagation Glass block Figure (b) (b) Figure (b) shows a plane light wavefront incident on a glass block of refractive index 1.50. The four parallel lines represent the positions of a wavefront at four equal time intervals. Draw the positions of the wavefront after AB for six equal time intervals. 2. (a) What is meant by a wavefront? Show on a diagram, the direction of a ray relative to a wavefront. (b) By considering the distribution of energy on a wavefront, derive a relationship between the amplitude of a wave and the distance r from (i) a point source of spherical wavefronts, (ii) a line source that produces cylindrical wavefronts. 23.2 Interference Students should be able to: • explain the concept of coherence • explain the concept of optical path difference, and solve related problems • state the conditions for constructive and destructive interferences Learning Outcomes 2013/P3/Q9, 2014/P3/Q9, 2015/P3/Q8

Physics Term 3 STPM Chapter 23 Wave Optics 161 23 Optical Path 1. Definition: The optical path of light in a medium is given by optical path = nl where n = refractive index of medium l = geometrical distance travelled by light in the medium 2. Figure 23.3 shows a plane wavefront AB incident on a glass surface in vacuum. If t = time taken for the wavelet from the point B to reach C, then BC = ct c = speed of light in vacuum The optical path in vacuum = n × BC = nct n = 1 for vacuum = ct ............................. ① 3. At the same time t, the wavelet from the point A has travelled a distance AD, where AD = vt v = speed of light in glass Optical path in glass = n × AD = nvt n = refractive index of glass But refractive index, n = c v v = c n Optical path in glass = n( c n ) t = ct ............................. ② Conclusion: From ① and ②, for the same time interval, the optical path is the same in all medium. l x O O A A n Vacuum Vacuum Medium Figure 23.4 4. Figure 23.4 (a) shows light travelling from O to A, in vacuum, a distance of l. Optical path in vacuum = l Figure 23.4 (b) shows a transparent material of thickness x and refractive index n placed between O and A. The total optical path from O to A = optical path in vacuum + optical path in medium = (l – x) + nx = l + (n – 1) x Hence the optical path has increases by (n – 1)x. This is due to light taking a longer time to travel from O to A. Light travels slower in the medium of refractive index n. A B D C Vacuum Glass Figure 23.3

Physics Term 3 STPM Chapter 23 Wave Optics 162 23 Example 2 A convex lens of diameter 4.0 cm has zero thickness at the edge. An object O is place on the principal axis of the lens at a distance of 50 cm from the optical centre of the lens. An image is formed on the opposite side of the lens. The image distance equals the object distance. The refractive index of the material of the lens is 1.5. If the number of wavelengths along the path of the light via the edge is the same as that along the principal axis, calculate the thickness of the lens at the centre of the lens. Solution: 50 cm 50 cm B x x 2 cm P A O I The fi gure above is symmetrical about the line AB. OA = OP2 + AP2 = 502 + 22 = 50.04 cm If λ0 cm = wavelength of light in air, then the number of wavelengths from O to A = 50.04 λ0 ............................................. ① If λ1 cm = wavelength of light in the lens, then the number of wavelengths from O to P = number of wavelength in air + number of wave in lens = (50 – x) λ0 + x λ1 .............................................. ② Given that ① = ② 50.04 λ0 = (50 – x) λ0 + x λ1 50.04 = (50 – x) + λ0 λ1 x ( λ0 λ1 = refractive index = 1.5) = 50 – x + 1.5 x 0.5x = 0.04 x = 0.08 cm Thickness of lens at the centre = 2x = 0.16 cm ABBBBBBB ABBBBBB

Physics Term 3 STPM Chapter 23 Wave Optics 163 23 Interference 2012/P2/Q3 1. Definition: Interference is the effect produced by the superposition of waves from two coherent sources. 2. The wavelength of light is in the range 400 nm to 700 nm which is rather small compared to wavelength of mechanical waves such as sound waves or water waves. 3. The conditions required for the formation of clear static interference pattern of light include: (a) The waves must be from two coherent light sources. Two sources of waves are coherent if the phase difference between the sources is constant, that is does not change with time. Hence the sources have the same frequency. (b) The waves have the same amplitude or at least about the same amplitudes. (c) The separation between the two coherent sources must be small, otherwise the interference pattern would not be seen clearly by the eye. (d) The distance of the screen or the observer from the two coherent sources must be sufficiently large. (e) The path difference of light from the two coherent sources at a point on the screen must be small. 4. Constructive interference occurs between two waves, y1 and y2 occurs when the amplitude of the resultant wave (y1 + y2 ) is greater than the amplitudes of y1 and y2 . 5. For maximum constructive interference at a point, the waves from the coherent sources, S1 and S2 arriving at the point must be in phase as shown by the displacement-time graphs of the waves (Figure 23.5). Maximum constructive interference • y1 and y2 are in phase • Amplitude of resultant wave (y1 + y2 ) is (A + A) = 2A y1 A −A t 0 y2 A −A t 0 y1 + y2 2A -2A t 0 Figure 23.5

Physics Term 3 STPM Chapter 23 Wave Optics 164 23 6. Figure 23.6 shows waves from two coherent sources arriving at a point, P in phase to produce maximum constructive interference. S1 and S2 emitting waves in phase. Waves arriving at P in phase. Path difference, S2 P – S1 P = 6 λ – 6 λ = 0 P S1 S2 S1 and S2 emitting waves antiphase. Waves arriving at P in phase. Path difference, S2 P – S1 P = 6 1 λ – 6 λ = 1 λ P 2 2 S1 S2 Figure 23.6 7. Hence for maximum constructive interference at a point, • If the coherent sources emit waves in phase, then path difference = 0, λ, 2λ, 3λ , … which is equivalent to phase difference = 0, 2π, 4π, 6π radians, … • If the coherent sources emit waves in antiphase. path difference = 1 2 λ, 11 2 λ, 21 2 λ, 31 2 λ,… which is equivalent to phase difference = π, 3π, 5π, 7π radians, … 8. Figure 23.7 shows waves from two coherent sources arriving at a point, P in antiphase to produce maximum destructive interference. S1 and S2 emitting waves in phase. Waves arriving at P in antiphase. Path difference, S2 P – S1 P = 6 1 λ – 6 λ = 1 λ P 2 2 S1 S2 S1 and S2 emitting waves in antiphase. P S1 S2 Waves arriving at P in antiphase. Path difference, S2 P – S1 P = 6 λ – 6 λ = 0 Figure 23.7

Physics Term 3 STPM Chapter 23 Wave Optics 165 23 9. Hence for maximum destructive interference at a point, • If the coherent sources emit waves in phase, then path difference = 1 2 λ, 11 2 λ, 21 2 λ, 31 2 λ,… = (m – 1 2)λ m = 1, 2, 3,… which is equivalent to phase difference = π, 3π, 5π, 7π radians,… = (m – 1 2)2π rad m = 1, 2, 3,… • If the coherent sources emit waves in antiphases, path difference = 0, λ, 2λ, 3λ,… = mλ m = 1, 2, 3,… which is equivalent to phase difference = 0, 2π, 4π, 6π radians,… = 2mπ rad m = 0, 1, 2, 3,… Quick Check 2 1. Light wave of wavelength travels from a point, P to a point, Q separated by distance x. A glass block of refractive index n and thickness d is placed between the points P and Q. (a) What is the change in the optical path? (b) What is the phase change at the point Q? 2. Discuss whether the following point sources S1 and S2 are coherent. (a) S1 and S2 have different frequencies but are of the same amplitude. (b) S1 and S2 have the same frequency, but are in antiphase. (c) S1 and S2 have the same frequency, but S1 lags behind S2 by π 2 radians. 3. Water waves of wavelength 2 cm are produced by two generators S1 and S2 separated by a distance of 4 cm. The points X, Y and Z are along a straight line with Y as the midpoint. X 4 cm 3 cm Y Z S2 S1 Discuss what type of interference occurs at the points X, Y and Z if S1 and S2 have the same frequency and are (a) in phase, (b) in antiphase. 23.3 Two-slit Interference Pattern Learning Outcomes Students should be able to: • explain Young’s two-slit interference pattern • derive and use the formula x = λD a for the fringe separation in Young’s interference pattern 1. In 1801 Thomas Young used a double slit to show interference of light. This experiment proved the wave theory of light. 2013/P3/Q10, Q19, 2013/P3/Q9, 2014/P3/Q13, 2016/P3/Q19, 2017/P3/Q16

Physics Term 3 STPM Chapter 23 Wave Optics 166 23 Monochromatic light source C A B a D Q P Bright Dark O Bright Dark Bright Screen Figure 23.8 2. Figure 23.8 shows the arrangement used in Young’s double-slit experiment. A monochromatic light source is placed behind a narrow slit C, which acts as a point source. 3. According to Huygens’s principle, spherical wavefronts from the slit C is incident on the double slit A and B which are parallel to slit C. 4. Since the slits A and B are equidistance from C, the same wavefront arrives at slits A and B simultaneous. Since points on the same wavefront act as secondary sources emitting wavelets in phase, spherical wavefronts are emitted in phase from slits A and B . Hence the slits A and B acts as coherent sources, having the same frequency and amplitude. 5. This method of producing two coherent sources A and B from a single source C is known as division of wavefront. Figure 23.9 6. Superposition of wavefronts from the double slit produce the static interference pattern on the screen. The interference pattern consists of parallel bright and dark fringes as shown in Figure 23.9. 7. Since the slits S1 and S2 are coherent sources that are in phase, the condition for (a) constructive interference, or bright fringe is optical path difference = mλ m = 0, 1, 2, 3,… (b) destructive interference, or dark fringe is optical path difference = (m – 1 2)λ m = 1, 2, 3,… 8. In Figure 23.8, AO = BO hence optical path difference = 0, and constructive interference produces a bright fringe at O. This bright fringe is known as the central maximum (m = 0).

Physics Term 3 STPM Chapter 23 Wave Optics 167 23 9. At the point Q is the first dark fringe from the central maximum, m = 1. optical path difference = (m – 1 2)λ = (1 – 1 2)λ = 1 2 λ 10. Hence in the equation, optical path difference = (m – 1 2)λ m = 1, 2, 3,… m represents the mth dark fringe from the central maximum. 11. The graph in Figure 23.10 shows how the intensity of light varies with the distance from the axis of the interference pattern for a few fringes near the axis. The maximum intensity at the centre of each bright fringe is almost the same for these few fringes. The minimum intensity is at the centre of each dark fringe and is zero. Derivation of the Formula x = λD a Monochromatic light source C A Q θ θ B N D P O xm –1 xm a Screen Interference pattern Figure 23.11 1. Figure 23.11 shows the arrangement in Young’s double-slit experiment, not drawn to scale. Suppose that the slit separation is a, the actual value of a is about 0.1 mm. The distance of the screen from the double slit is D which is about 50 cm. The distance of P the mth bright fringe from O, the centre of the interference pattern is xn . 2. N is the point on the line PB such that PN = PA Since the slit separation a ( 0.1 mm) is small compared to D ( 50 cm) the angle APN is rather small (<<1°). Hence angle ANP is almost 90°. Also angle PQO = angle BAN = θ Dark fringe Dark Dark Dark Dark Bright Bright Bright Bright Bright Dark Distance Intensity O O Figure 23.10

Physics Term 3 STPM Chapter 23 Wave Optics 168 23 3. At the point P, the optical path difference of waves from the slits A and B is given by optical path difference = PB – PA = PB – PN (PA = PN) = BN In the right-angled triangle ANB, BN = a sin θ ............................. ① Since P is the position of the mth bright fringe from O, optical path difference = mλ ............................. ② From ① and ②, a sin θ = mλ m = 0, 1, 2, ... 4. In the triangle POQ, the angle θ is small, hence sin θ = tan θ = xn D From a sin θ = mλ a ( xm D ) = mλ Distance of mth bright fringe from O xm = mλD a m = 0, 1, 2, .... For the (m – 1)th bright fringe from O, xm – 1 = (m – 1)λD a 5. The separation between two neighbouring bright fringes or two neighbouring dark fringes is known as the fringe width x. Fringe width, x = xm – xm – 1 = mλD a – (m – 1)λD a x = λD a 6. Similarly the distance of the mth dark fringe from the point O, xm = m – 1 2 λD a m = 1, 2, 3, ... Example 3 In a Young’s double-slit experiment, the slit separation is 0.05 cm and the distance of the double slit from the screen is 200 cm. When blue light is used, the distance of the fi rst bright fringe from the centre of the interference pattern is 0.13 cm. (a) Calculate the wavelength of the blue light. (b) Calculate the distance of the fourth dark fringe from the centre of the interference pattern. (c) The space between the double slit and the screen is fi lled with water. Explain the changes to the interference pattern.

Physics Term 3 STPM Chapter 23 Wave Optics 169 23 Solution: (a) Slit separation, a = 0.05 cm = 5.0 × 10–4 m Distance of screen from double slit, D = 200 cm = 2.00 m Fringe width, x = 0.13 cm = 1.3 × 10–3 m Using x = λD a Wavelength, λ = xa D = (1.3 × 10–3) (5.0 × 10–4) 2.00 = 3.25 × 10–7 m (b) x x x x O 1st 2nd 3rd 4th dark fringe Centre 1 – 2 From the fi gure above, distance of 4th dark fringe from centre of interference pattern O = 3.5 x = 3.5 × 0.13 cm = 0.455 cm (c) When water fi lls the space between the double slit and screen, λ, the wavelength of light in water is shorter. From the equation x = λD a , the fringe width x decreases. Quick Check 3 1. S S1 S2 a b g f Screen The fi gure represents a Young’s double-slit experiment, a is the width of the single slit S, b the distance of the double-slit S1 S2 from S, f the slit separation of the double-slit and Δx the fringe separation on the screen. If λ is the wavelength of the monochromatic light used, then the fringe separation Δx is A λf g C λb g B λg f D λb a 2. Coherent light waves emerge from two fi ne parallel slits S1 and S2 as shown in the fi gure. P S1 S2 A dark fringe occurs at P. If m is an integer, the phase difference between the wavefronts from S1 and S2 at P is A 2mπ rad. C (2m + 1 2 ) π rad. B (m + 1 2 ) π rad. D (2m + 1)π rad.

Physics Term 3 STPM Chapter 23 Wave Optics 170 23 3. Two identical narrow slits S1 and S2 are illuminated by light of wavelength λ from a narrow slit P. P Q l1 S1 S2 l2 l3 l4 The light from S1 and S2 falls on a screen. If m is an integer, the condition for destructive interference at Q is that A (l 1 – l 2 ) = (2m + 1) λ 2 B (l 3 – l 4 ) = (2m + 1) λ 2 C (l 1 + l 3 ) – (l 2 + l 4 ) = (2m + 1) λ 2 D (l 1 + l 3 ) – (l 2 – l 4 ) = mλ 4. Two coherent monochromatic waves of equal amplitude are brought together to interfere on a screen. Which one of the following graphs correctly represent the variation of intensity with distance x across the interference pattern? A 0 x Intensity B Intensity 0 x C Intensity 0 x D Intensity 0 x 5. In a Young’s double-slit experiment, the intensity of light at the centre of the fringe pattern is I. If one of the two identical slits is now covered, the intensity at the centre is A I C I 2 B AB I 2 D I 4 6. Water waves of wavelength 2 cm are produced by two generators, S1 and S2, placed 4 cm apart. Each generator, when operated by itself, produces waves which have amplitude A at the point P which is 3 cm from S1 as shown in the figure. 4 cm 3 cm S1 S2 P If the generators oscillate in phase, the amplitude of oscillation at P is A 0 C 2A B 1 2 A D 4A 7. In a Young’s double-slit experiment, light of wavelength 600 nm produces fringes of separation 0.4 mm on a certain screen. With a different source of monochromatic light, the width of 10 bright fringes on the same screen is 3.3 mm. What is the wavelength of the second light source? A 495 nm C 727 nm B 660 nm D 990 nm 8. Which of the followings is not necessary for the superposition of two waves to produce interference at a point? A The waves must have the same wavelength. B The phase difference between the waves must be constant. C The waves must have the same amplitude. D The waves must be transverse waves. 9. In a Young’s two-slit experiment, monochromatic light of wavelength 500 nm are used and the separation between the slits is 0.25 mm. What is the angular separation between the second-order maximum? A 2.0 × 10–3 rad C 6.0 × 10-3 rad B 4.0 × 10–3 rad D 8.0 × 10–3 rad

Physics Term 3 STPM Chapter 23 Wave Optics 171 23 Changes in the Interference Pattern 2009/P1/Q42, 2010/P1/Q40 1. The fringe separation x in Young’s double-slit experiment is given by x = λD a The variables that determine the fringe separation are (a) λ, the wavelength of light (b) a, the slit separation (c) D, the distance of the screen from the double slit Hence when anyone of these variables changes, we can explain any change to the fringe pattern by referring to the equation x = λD a 2. If the slit separation a is bigger, but λ and D remains unchanged, from the equation x = λD a the fringe separation x decreases. 3. If the screen is moved further from the double slit, D increases, from x = λD a the fringe separation x increases. The maximum intensity of the bright fringes decreases. 4. Experiment using blue light would produce fringe pattern of smaller fringe separation compared with experiment using the same arrangement but a monochromatic red light source is used instead. The wavelength of blue light is shorter than that of red light. 5. Besides the three variables a, λ and D that affects the interference pattern, changes to the interference pattern are also caused by (a) the distance of the monochromatic light source from double slit. If the single slit together with the light source is more closer to the double slit, intensity of light from the double slit increases. Hence the bright fringes of the interference pattern would be brighter. (b) filling the space between the double slit and the screen with a transparent medium. The wavelength of light in a medium of refractive index μ is λ μ, where λ = wavelength of light in air. Hence the wavelength is decreases and from x = λD a , the fringe separation decreases from x to x µ . (c) substituting the monochromatic light source with a white light source. White light consists of various colours red, orange, yellow, ...., violet of different wavelengths. The wavelength of red light is the longest and that of violet light is the shortest. Hence the interference pattern would consists of bright fringes of various colours each of different fringe separation. Hence the fringes are not separated uniform. Since all wavelengths produce constructive interference at the axis of the interference pattern, the central maximum is a white fringe.

Physics Term 3 STPM Chapter 23 Wave Optics 172 23 (d) covering one of the two slits with a thin transparent film as shown in Figure 23.12. C A B d Thin film Screen O O′ Figure 23.12 The central maximum is shifted from O to O’. This is noticeable when white light is used because the central maximum is white whereas the other bright fringes are coloured. Explanation: Optical path of light from slit A to O = AO Optical path of light from slit B to O = BO + (µ – 1)d = AO + (µ – 1)d (BO = AO) when µ = refractive index of material of transparent film. Hence the optical path difference of light from A and from B = [AO + (µ – 1)d] – AO = (µ – 1) d ≠ 0 Hence the point O is not the central maximum (m = 0). Constructive interference or destructive interference may occur at O. If constructive interference occurs, a bright fringe is formed at O, and optical path difference = mλ (μ – 1)d = mλ where m is the number of bright fringe from O’ to O. If destructive interference occurs, a dark fringe is formed at O, and optical path difference = (m – 1 2 ) λ (μ – 1)d = (m – 1 2 ) λ where m is the number of dark fringe from O′ to O. (e) placing a piece of polaroid in front of each slit of the double slit. Sharp interference fringes would be produced on the screen if the axis of polarisation of the polaroids are parallel. When one of the polaroids is rotated through an angle θ (θ ≠ 90°), the planes of polarisation of the plane-polarised light from the slits are inclined at an angle θ to each other. The variation of intensity with distance from the central maximum θ is as shown in Figure 23.13. Intensity Distance Figure 23.13

Physics Term 3 STPM Chapter 23 Wave Optics 173 23 The intensity of the bright fringe decreases, and the intensity is no more zero when destructive interference occur. Hence the difference of intensity between the bright and ‘dark’ fringes decrease. When the plane of polarisation of the polaroids are perpendicular to each other no interference occurs. The intensity of light on the screen is uniform. (f) covering one of the slits with an opaque card. No more interference pattern is produced on the screen. Instead, single slit diffraction pattern as shown in Figure 23.14 is obtained. Figure 23.14 (g) substituting the double slits with two bulbs emitting monochromatic light of the same frequency. No interference pattern is formed on the screen because the two bulbs are not coherent sources. Wavetrains are emitted randomly from each bulb. The phase difference is not constant. Example 4 In a Young’s double-slit experiment, the slit separation is 0.20 mm and the distance of the screen from the double slit is 1.00 m. (a) If the wavelength of light used is 644 nm, calculate the angle separation between two neighbouring dark fringes on the screen when viewed from the double slit. (b) When the double slit is substituted with another, the fringes on the screen are just able to be resolved by the eye at the position of the double slit. (i) If the resolving power of the eye is 3.0 × 10–4 radians, what is the slit separation of the double slit? (ii) Using the double slit in (b), explain what modifi cation to the experimental set up that would enable the fringes on the screen to be seen clearly again. Give a quantitative example for your answer. If the screen is then moved further from the double slit, describe the change to the fringe pattern. Solution: (a) Using x = λD a Angular separation of fringes θ = x D = λ a = 644 × 10–9 0.20 × 10–3 = 3.22 × 10–3 rad. D x θ

Physics Term 3 STPM Chapter 23 Wave Optics 174 23 (b) (i) When the eye is just able to resolve the fringes, angular separation, θ = resolving power of eye λ a = 3.0 × 10–4 slit separation, a = 644 × 10–9 3.0 × 10–4 = 2.15 × 10–3 m = 2.15 mm (ii) In order to see the fringes on the screen clearly, the angular separation θ must be increased. Since θ = λ a to increase θ with slit separation a is fi xed, use light of longer wavelength, for example use λ = 700 nm. When the distance D of the screen from the double slit is increased, 1. from the equation x = λD a , the fringe separation x increases. 2. the intensity of the bright fringes decreases. Example 5 A Young’s double-slit system is fi xed to one wall of an empty rectangular glass tank. A sodium vapour lamp with a narrow slit is arranged behind the double slit so that interference fringes are produced on the opposite wall of the glass tank. When the tank is fi lled with a type of oil, the fringe separation changes by 35 %. (a) Explain why, and discuss whether the fringe separation increases or decreases. (b) Calculate the refractive index of the oil. Solution: (a) Fringe separation, x = λD a ............................. ① In oil of refractive index µ, the wavelength of light decrease because λ’ = λ µ (µ > 1) < λ Hence the fringe separation decreases because x’ = λ’D a = λD µa ............................. ② < x (b) ① ②: x’ x = 1 µ , x’ = 65% x = 0.65x 0.65x x = 1 µ Refractive index, µ = 1 0.65 = 1.54

Physics Term 3 STPM Chapter 23 Wave Optics 175 23 Example 6 In a Young’s double-slit experiment, monochromatic light of wavelength 500 nm produced an interference pattern on the screen. When a thin fi lm of transparent material of refractive index 1.5 covers one of the slits, the central maximum is displaced to the position of the ninth bright fringe. Calculate the thickness of the fi lm. Solution: The fi gure shows the central maximum displaced from O to O’ the position of the 9th bright fringe in the original interference pattern. A O′ O B d Hence at O, along the axis of the system is the 9th bright fringe. Optical path difference = mλ m = 9 AO + (µ – 1)d – BO = 9λ AO = BO (µ – 1) d = 9λ Thickness of fi lm, d = 9 × (500 × 10–9) (1.50 – 1) = 9.0 × 10–6 m Example 7 In a Young’s double-slit experiment, a light source that emits two different wavelengths of 7 × 10–7 m (red) and 5.6 × 10–7 m (green) is used. (a) Draw a sketch to show the interference fringes for both the colours. (b) Show that when fringes of the two colours exactly overlap for the fi rst time from the centre of the fringe pattern, (m + 1) m = 5 4 . What is the signifi cant of m? What is the colour of the fringe at that point? Solution: (a) For red light, λ1 = 7.0 × 10–7 m For green light, λ2 = 5.6 × 10–7 m Fringe separation, x1 = λ1 D a Fringe separation, x2 = (5.6 × 10–7) D a = (7.0 × 10–7) D a x2 x1 = 5.6 7.0 = 0.8 x2 = 0.8 x

Physics Term 3 STPM Chapter 23 Wave Optics 176 23 The fringe patterns (drawn to scale) on one side of the central maximum are shown below. (i) Red fringes (ii) Green fringes (iii) Combined fringe pattern (b) When a red fringe and a green fringe overlap, m = 4 for red and m = 5 for green (fi gure above) = (4 + 1) Hence the mth red fringe from the centre overlaps with the (m + 1)th green fringe. Hence mx1 = (m + 1)x2 m(7.0 × 10–7) D a = (m + 1) (5.6 × 10–7) D a m + 1 m = 7.0 5.6 = 5 4 m = 4 Hence m represent the fourth red fringe overlaps with the fi fth green fringe. The colour of fringe where overlapping occurs is (red + green) = yellow. Experimental Determination of the Wavelength of Light S C A B D Converging lens Monochromatic light source Interference fringes Eye Eyepiece a Figure 23.15 1. Figure 23.15 shows the Young’s double-slit experiment used to determine the wavelength λ of a monochromatic light source. R R R R R R G G G G G G G G G R G R G R G G +R G R m = 0 m = 1 m = 2 m = 3 m = 4 m = 5 m = 0 m = 1 m = 2 m = 3 m = 4 m = 5 m = 6 x2 = 0.8x1 Overlap

Physics Term 3 STPM Chapter 23 Wave Optics 177 23 A converging lens is used to focus light from the narrow slit through the slit C in order to produce a brighter fringe pattern. 2. An eyepiece fi tted with a cross-wire acts as a magnifying glass is used to observe the fringe pattern. A vernier scale fi tted beside the eyepiece enables the fringe separation to be measured. 3. The eyepiece is fi rst moved until the cross-wire is at the position of a bright fringe. The initial reading l 0 of the vernier scale is noted. The eyepiece with the cross-wire is then slowly moved across 10 bright fringes and the fi nal scale reading l 1 noted. Hence l 1 – l 0 = 10x x = 1 1.0 (l 1 – l 0 ) 4. Using λ = xa D the wavelength λ of the light can be calculated. Quick Check 4 1. In the Young’s double-slit arrangement shown in the fi gure below, a pattern of equally spaced parallel fringes appears on a screen S. Which one of the following quantities, if increased, would cause the fringe separation to increase? z x y d S y l A x C d B y D l 2. Coherent 3 cm-wave sources S and S′ are placed a distance d apart as shown in the fi gure. Y Y' L d S S' What are suitable values for d and L if interference fringes are to be formed at YY ′ ? d/mm L/mm A 1 1 000 B 3 100 C 30 1 000 D 300 1 000 3. Under which of the following sets of conditions will the fringe separation of a double slit interference pattern be greatest? Slit Distance of slits Wavelength separation from screen A small small short B small large long C large small long D large large short 4. The fi gure below shows the arrangement that produces interference fringes with a double slit. Thin glass plate S1 S2 X O Y Monochromatic ∗ source When slit S2 is covered with a very thin plate of glass

Physics Term 3 STPM Chapter 23 Wave Optics 178 23 A the fringe separation increases B the fringe separation decreases C the fringe pattern moves towards X D the fringe pattern moves towards Y 5. A student who carried out Young’s doubleslit experiment could not observe any fringe pattern but found that the screen to be uniformly lighted. Which one of the following procedure would enable the fringe pattern to be seen? A Increase the intensity of light source. B Move the screen closer to the double slit. C Decrease the separation between the slits. D Use light of shorter wavelength. 6. A very thin piece of polaroid is used to cover the slit S1 in a Young’s double-slit experiment as shown in the figure. P O Q Screen S1 S2 Monochromatic ∗ source Thin Polaroid Which of the following describes the changes to the original interference pattern produced on the screen? A The intensity of the bright fringes decreases and the central maximum shifts towards P. B The intensity of the bright fringes increases and the central maximum shifts towards Q. C The intensity of the bright fringes decreases and the fringe separation increases. D The intensity of the bright fringes increases and the fringe separation decreases. 7. In a Young’s double-slit experiment, a very thin transparent material of refractive index 1.58 covers one of the slits. The centre of the screen is now occupied by the seventh bright fringe of the interference pattern. If the wavelength of light used is 550 nm, calculate the thickness of the thin material. 8. (a) State the conditions necessary to produce sharp interference pattern from two sources of light. (b) Draw a figure to show the arrangement used in Young’s double-slit experiment to measure the wavelength of light. (c) In one such experiment, the following readings were recorded. Separation between the slits = 0.11 cm Distance of double slit from the plane where the fringes are observed = 98.2 cm Distance from the nth bright fringe to the (n + 10)th bright fringe = 0.52 cm Calculate the wavelength of light. 9. In a Young’s double-slit experiment, the separation between yellow fringes is 0.275 mm. The yellow light of wavelength 550 nm is then replaced by magenta light which consists of a combination of violet light of wavelength 400 nm and red light of wavelength 600 nm. The arrangement of the experiment remain unchanged. Calculate (a) the fringe separation for violet light, (b) the fringe separation for red light, (c) the separation between the magenta fringe on the axis and next magenta fringe. Hence, sketch a diagram to show the fringe pattern formed using magenta light, including the colours and extend to 1 mm from the axes. 10. (a) Explain what is meant by (i) coherent, and (ii) interference. (b) S ∗ A B C d D Screen The figure shows the arrangement in a Young’s double-slit experiment. (i) Write an expression relating the fringe separation s on the screen with the wavelength λ, the slit separation d, and the distance of the screen from the double slit, D. (ii) Suggest suitable values for d, and D. (c) Describe and explain the changes to the fringe pattern if (i) both the slits B and Care narrower, but the slit separation d remains unchanged,

Physics Term 3 STPM Chapter 23 Wave Optics 179 23 (ii) the intensity of light from the slit B is halved the intensity of light from slit C, (iii) a thin piece of transparent plastic is used to cover the slit B, (iv) each of the slits B and C is covered by a piece of polaroid and the polaroid in front B is slowly rotated. 11. (a) State two conditions for two sources of light to produce interference pattern that is observable. (b) (i) In Diagram (i) the source S that emits waves of wavelength λ and is equidistance from the slits X and Y. The point P is along the axis of the two-slit. Discuss whether constructive or destructive interference occurs at the point P. (i) X Y S P (ii) X Y S P (ii) The wavelength λ = 600 nm, and separation between the slits is 0.500 mm. Interference fringes are formed on a screen 0.80 m from the two-slit. Determine the distance of the fourth bright fringe on the screen from the centre of the interference pattern. (c) In Diagram (ii), the source S is not on the axis of the two-slit, and (SX - SY) = λ. The point P is along the axis of the two-slit. Discuss whether constructive or destructive interference occurs at the point P. 23.4 Interference in a Thin Film Learning Outcome Students should be able to: • explain the phenomenon of thin fi lm interference for normal incident light, and solve related problems 1. Colours seen in a soap bubble, a thin soap fi lm and in the layer of oil fl oating on water are due to interference of light. 2. Figure 23.16 shows monochromatic light at almost normal incidence to a thin film of thickness t and refractive index µ. The light is partially refl ected at A along AX and partially refracted into the fi lm and then refl ected at B along CY. 3. The rays AX and CY are coherent. There is a phase change of π radian which is equivalent to a path difference of λ 2 due to refl ection at a denser medium at A. Whereas no phase difference for refl ection at B. 4. If t = thickness of fi lm. Total optical path difference between the rays AX and CY = 2μt – λ 2 If (2 µt – λ 2 ) = mλ m = 0, 1, 2, 3, ... constructive interference occurs and a bright fringe is observed. On the other hand, if (2 µt – λ 2 ) = (m – 1 2 ) λ 2µt = mλ m = 1, 2, 3, ... destructive interference occurs and a dark fringe is observed. X Y O B Monochromatic light Thin film t A C Figure 23.16 2008/P1/Q41, 2015/P3/Q9, 2017/P3/Q8

Physics Term 3 STPM Chapter 23 Wave Optics 180 23 Film t P B Q X Y A C Figure 23.17 5. Interference pattern can also be observed in the light transmitted by a thin film as shown in Figure 23.17. Part of the light transmitted is refracted into air along AX, partly reflected internally along AB, reflected at B and finally emerging into air. The rays X and Y that enter the eye are coherent and produce interference when focussed by the lens of the eye. 6. Internal reflection at A and B does not cause any phase change. Hence total optical path difference between rays X and Y is 2µt. Constructive interference occurs if, 2µt = mλ m = 1, 2, 3, ... and a bright fringe is observed. Destructive interference occurs, if 2µt = (m – 1 2 ) λ m = 1, 2, 3, ... and a dark fringe is observed This is opposite to the interference produced by reflected light discussed in (4) above. 7. An application of interference in thin film is the blooming of optical lenses to reduce reflection of light from lenses of spectacles. X A Y t B Lens Air Film Figure 23.18 8. The surface of the lens is coated with a thin film so that destructive interference occurs between the light reflected from the surface of the film and that reflected from the lens surface. 9. There is a phase change of π radian which is equivalent to a path difference λ 2 when light is reflected at A on the surface of the film. The refractive index µ of the film is lower than the refractive index of glass. Reflection at B also causing a path difference of λ 2 . Hence reflection at A and B does not produce any difference in optical path. The path difference between the reflected rays X and Y is only 2µt.

Physics Term 3 STPM Chapter 23 Wave Optics 181 23 10. The minimum thickness t of the fi lm which would produce destructive interference is given by 2µt = λ 2 thickness of fi lm, t = λ 4 µ 11. Since white light consists of different wavelengths, destructive interference does not occur simultaneous for all the different wavelengths. For blooming of lenses, a wavelength which equals the average wavelength of white light is chosen and substituted into the equation t = λ 4µ . This average wavelength is the wavelength of yellow-green light. Destructive interference occurs for yellow-green light leaving the red and blue components of white to be refl ected. Since red + blue = magenta, the lens has a magenta hue. Info Physics Non-refl ective lenses of different colours are available. The colour of a non-refl ective lens depends on the value of the wavelength of light that is used to fi x the thickness of the fi lm – recall the equation t = λ 4µ . If the value of λ chosen is 550 nm, that of green-yellow, then the colours of light refl ected would mainly be the red and blue end of the spectrum. The combined colour is magenta. Example 8 Light of wavelength 5.9 × 10–7 m falls almost normally on a vertical soap fi lm. The refl ected light produces parallel, horizontal fringes separated uniformly with fringe separation 4.0 mm. What can be deduced about the soap fi lm? (Refractive index of soap fi lm = 1.34) Solution: Since the fringes are parallel, horizontal and separated evenly, the soap fi lm forms a wedge as shown in the fi gure. If t = thickness of wedge, where a dark fringe is observed, then 2µt = mλ m = integer If x = fringe separation, then l = mx m = l x 2µt = mλ = lλ x Angle θ = t l = λ 2µx = 5.9 × 10–7 2(1.34)(4.0 × 10–3) = 5.50 × 10–5 rad. θ t l Monochromatic light Wedge of soap film Reflected light Fringes Light

Physics Term 3 STPM Chapter 23 Wave Optics 182 23 Example 9 Oil Water The fi gure shows an oil drop of refractive index µ = 1.20 fl oating on the surface of water of refractive index 1.33. Light refl ected from the oil drop is observed from above. (a) Explain whether a bright fringe or a dark fringe is observe at the edge of the oil drop. (b) Blue light of wavelength 450 nm is incident normally on the oil drop. Estimate the thickness of the oil drop at the position of the third dark fringe from the edge. (c) Explain why interference fringes are not seen when the thickness of the oil drop increases. Solution: Oil A B Air t (a) The fi gure shows light refl ected at A and B. There is a phase change which is equivalent to a path difference of λ 2 at A. At B, refractive index of oil, 1.20 < 1.33, refractive index of water. Hence refl ection at B also cause a path difference of λ 2 . Therefore the phase change due to refl ection at A and B does not produce any net path difference. If t = thickness of oil drop optical path difference between the two refl ected rays = 2µt At the edge of oil drop, t = 0. optical path difference = 0 Hence a bright fringe is observed. (b) When a dark fringe is formed, optical path difference = (m – 1 2 ) λ m = 1, 2, 3, ... 2µt = (m – 1 2 ) λ when m = 3, t = 3 – 1 2 (450 × 10–9) 2 × 1.20 = 470 nm (c) As the thickness of the oil drop does not increase uniformly, the fringes are not uniformly space. When the thickness of the oil drop increases, the separation between neighbouring fringes decreases until the eyes are not able to resolve them.

Physics Term 3 STPM Chapter 23 Wave Optics 183 23 Quick Check 5 1. When sunlight is refl ected from a wet road surface covered with a thin layer of oil, colourful pattern can be seen. These colours are due to A the natural colours of oil can be visible if the oil fi lm is very thin B selected wavelengths are obsorbed by the oil and the rest of the wavelengths refl ected C the phase difference between light reflected from the top surface and bottom surface of the oil film is dependent on the wavelength D small oil droplets are formed on the water surface and an effect identical to the formation of rainbow produces the different colours 2. The surface of a lens can be made nonrefl ective by coating the lens surface with a thin layer of transparent material. If λ is the wavelength of the incident light in the medium, the minimum thickness of the material required is A λ 8 C λ 8 B λ 4 D λ 3. A material of refractive index 3.50 is coated with a translucent fi lm of refractive index 2.00 as shown in the fi gure. n1 = 2.00 n2 = 3.50 Monochromatic light What is the minimum thickness of the fi lm so that light of wavelength 600 nm is not refl ected? A 45.0 nm B 75.0 nm C 90.0 nm D 150 nm 4. The fi gure shows a ray incident on a glass surface coated with a fi lm. n1 X Y n2 Air t Film Glass If the wavelength of light in the fi lm is λ, and the refractive indices of the fi lm and glass are n1 and n2 respectively, which of the following is the condition for destructive interference between the refl ected rays X and Y ? A t = λ 4 when n1 > n2 B t = λ 2 when n1 > n2 C t = λ 2 when n1 < n2 D t = λ when n1 < n2 5. A thin film of magnesium fluoride is deposited on the surface of a lens to make the lens non-refl ective. Discuss the factors that determine the thickness of the fi lm. Calculate a suitable value for the thickness of the fi lm. (Refractive index of magnesium fl uoride = 1.40, of glass 1.50, frequency of green light = 6.0 × 1014 Hz) 6. A thin fi lm of thickness 4.0 × 10–5 cm is illuminated almost normally using white light. If the refractive index of the fi lm is 1.50, calculate the wavelength of high intensity in the refl ected light.

Physics Term 3 STPM Chapter 23 Wave Optics 184 23 23.5 Diffraction by a Single Slit Learning Outcomes Students should be able to: • explain the diffraction pattern for a single slit • use the formula sin θ = λ a for the fi rst minimum in the diffraction pattern for a single slit • use the formula sin θ = λ a as the resolving power of an aperture 1. Besides interference, another wave property of light is diffraction. 2. Diffraction is the spreading of wave after a wave passing a narrow slit. 3. You can observe diffraction of light by viewing a bright source through the slit between two fi ngers held in front of your eye. Monochromatic light Slit Figure 23.19 4. Figure 23.19 shows the diffraction pattern produced on a plane in front of narrow single slit illuminated with monochromatic light. A X Y B O Screen Q Figure 23.20 5. Figure 23.20 shows plane wavefronts of light incident normally on a narrow slit AB. 6. According to Huygens’s principle, points on the plane wavefront AB emit wavelets in phase. 7. The optical paths of the wavelets from the points A and B to the point O on the axis of the slit are equal. Constructive interference occurs at O. Similarly, wavelets from points such as X and Y, such that QX = QY, also arrive at O in phase. Hence a maximum, or bright fringe is produced at O. 2015/P3/Q15, 2016/P3/Q11

Physics Term 3 STPM Chapter 23 Wave Optics 185 23 A X Q Y B Screen a P' O P θ C M R D S Figure 23.21 8. Figure 23.21 shows how the first minimum (dark fringe) is produced at the point P. The slit AB is divided into two zones AQ and QB. Rays AC and QD travelling towards P are inclined at an angle θ to the axis. When the path difference between the rays AC and QD which equals AM is λ 2 , destructive interference is produced at P. Similarly for rays from pairs of points such as X and Y (XY = a 2 ), the path difference is also λ 2 . Hence a dark fringe is produced at P. 9. In triangle AMQ, sin ∠AQM = AM AQ (∠AQM = θ) sin θ = λ 2 a 2 (AM = λ 2 , AQ = a 2 ) hence, sin θ = λ a is the condition of the first minimum. If θ is small, sin θ = θ hence, θ = λ a This equation gives us the angle θ above and below the central axis. It illustrates the extend that spread (diffracted) beyond the slit. 10. The positions of the other minima is given by sin θm = mλ a m = 1, 2, 3, … If θm is small, then θm = mλ a m = 1, 2, 3, … Exam Tips Dark fringes or minima are produced where the path difference between the top and bottom rays are equal to λ, 2λ, 3λ, … AM = λ, 2λ, 3λ,… M A B

Physics Term 3 STPM Chapter 23 Wave Optics 186 23 Intensity –3λ a 0 θ / rad –2λ a 2λ a 3λ a –λ a λ a Io Figure 23.22 11. Figure 23.22 shows the variation of the intensity of light across the single slit diffraction pattern when the width of the slit is a. 12. When the width of the slit is halved, from the equation: θ = λ a the angular separation between the first minimum and the central maximum is double. Whereas the intensity of the central maximum decreases to I0 4 and the other maxima are lower because less light passes through the narrow slit (Figure 23.23). Intensity 2 λ a 0 1 4 / rad Io broader θ –2 λ a Slit width = a 2 Figure 23.23 13. On the other hand, when the width of the slit is doubled, i.e. 2a, then the intensity of the central maximum is 4I0 , but the central maximum is narrower. (Figure 23.24). Intensity – λ –– 2a λ–– 2a 0 / rad 4 Io Slit width = 2a θ Figure 23.24

Physics Term 3 STPM Chapter 23 Wave Optics 187 23 14. To obtain sharp and bright single slit diffraction pattern, suffi cient intense light must pass through the slit. This is achieved using the arrangement of converging lenses shown in Figure 23.25. x O P L1 L f1 f θ Monochromatic S source Screen First minimum Figure 23.25 15. A converging lens L1 of focal length f 1 is placed at distance f 1 from a monochromatic light source S, so that rays emerging from the lens are parallel to the principle axis and incident normally on the slit. 16. Another lens L of focal length f is arranged in front of the slit to focus the diffraction pattern on the screen placed at a distance f from the lens L. 17. If P is the position of the fi rst minimum, then θ = λ a a = width of slit From Figure 23.26, θ = x f x = distance of P from O x f = λ a x = ( λ a ) f Example 10 In a single slit diffraction pattern, the distance between the fi rst minima on either sides of the central maximum is 5.2 mm. The screen is at a distance of 80.0 cm from the slit, and the wavelength of light is 546 nm. What is the width of the slit? Solution: Distance between the fi rst minimum and the central maximum x = 5.2 2 mm = 2.6 × 10–3 m Using θ = λ a and θ = x D λ a = x D Width of slit, a = λD x = (546 × 10–9)(0.80) (2.6 × 10–3) = 1.68 × 10–4 m

Physics Term 3 STPM Chapter 23 Wave Optics 188 23 Example 11 What is the angular separation between the fi rst minimum and the central maximum in the single slit diffraction pattern if the width of the slit is equal to (i) the wavelength of light? (ii) fi ve times the wavelength of light? (iii) ten times the wavelength of light? Solution: Using sin θ = λ a (i) When a = λ, sin θ = λ λ = 1 θ = 90° (ii) When a = 5λ, sin θ = λ 5 λ = 0.20 θ = 11°32’ (iii) When a = 10λ , sin θ = λ 10 λ = 0.10 θ = 5° 44’ Resolving Power 1. The resolving power of an optical instrument such as microscope, telescope and even the eye is its ability to distinguish between objects that are closely spaced. 2. With the use of a telescope, two distant stars that are close together can be distinguish. A microscope enables us to see cells that the naked eye is unable to see. 3. The resolving power of an optical instrument is limited by the diffraction of light. θ Q P θ Y X Figure 23.26 4. Figure 23.26 (a) shows the sources P and Q of light far from a single slit. The angle subtended at the slit, θ is large enough for the diffraction patterns formed on a screen to be distinguishable. The images of P and Q are said to be resolved.

Physics Term 3 STPM Chapter 23 Wave Optics 189 23 5. In Figure 23.26 (b) X and Y are also sources of light far from the single slit. The angle subtended at the slit, θ is too small for the diffraction patterns to be distinguishable. The two central maxima cannot be detected. The images are not resolved. 6. Rayleigh’s criterion states that for the images to be just resolved the central maximum of one image falls on the fi rst minimum of the other image (Figure 23.27). First minimum First minimum Resultant intensity Figure 23.27 Rayleigh’s criterion 7. Hence for the two images to be resolved, the minimum angle θmin subtended by the sources at the slit equals the angular separation between the central maximum and the fi rst minimum of the single-slit diffraction pattern. sin θmin = λ a where λ is the wavelength and a is the width of the slit. This ratio sin θmin = λ a is known as the resolving power of the aperture. 8. The wavelength λ of light is very small compared to the width of the slit a (λ << a). Since θmin is small, and sin θmin ≈ θmin in radians. Hence resolving power, θmin = λ a 9. Most optical instruments have circular aperture. The resolving power for a circular aperture of diameter a is given as θmin = 1.22 λ a Info Physics Electron microscope has very high resolving power, the angle θmin, is very much smaller than that of optical microscope. Electron microscope uses a beam of electrons to ‘illuminate’ the specimen. The wavelength of an electron beam is 1/104 , the wavelength of light. Hence the value of θmin for an electron microscope is 1/104 that of an optical microscope. Example 12 The diameter of the pupil of a person’s eye is 2 mm. The average wavelength of light is 550 nm. (a) Estimate the resolving power of the eye. (b) Two parallel lines that are separated by a distance of 1 cm are drawn on a piece of card. What is the maximum distance of the card from the eye for the images of the lines to be resolved?

Physics Term 3 STPM Chapter 23 Wave Optics 190 23 Solution: (a) The pupil of the eye is circular, resolving power of the eye, θmin = 1.22 λ a = 1.22 ( 550 × 10–9 2 × 10–3 ) rad = 3 × 10–4 rad (b) 1 cm D Eye θ min Since θmin = 3 × 10–4 rad is small, θmin = 3 × 10–4 rad = 0.01 D Maximum distance, D = 0.01 3 × 10–4 m = 30 m Quick Check 6 1. Monochromatic light of wavelength λ is at normal incidence on a single slit of width a. The distribution of the intensity across the screen is as shown below. Intensity 0 sin θ a – a I0 λ λ If the slit is replaced with one of width 2a, which is the correct intensity distribution graph? A 0 Intensity sin θ 2a – 2a 2I0 λ λ C 0 Intensity sin θ 2a – 2a 4I0 λ λ B 0 Intensity 2 sin θ a – 2 a 2I0 λ λ D 0 Intensity sin θ 2a – 2a 4I0 λ λ 2. A beam of laser is directed at right angles to a single narrow slit. Which combination of wavelength and slit width causes the width of the central maximum formed on a screen to be smaller. Wavelength Slit width A Longer Wider B Shorter Wider C Longer Narrower D Shorter Narrower 3. Monochromatic light incident on an adjustable single slit produced on a screen a diffraction pattern. The fi gure shows the variation of intensity with distance across the pattern. Io Intensity Central maximum Subsidiary maxima 0 Distance When the width of the slit was reduced, A the intensities of all the maxima increased and the width of the pattern decreased

Physics Term 3 STPM Chapter 23 Wave Optics 191 23 B the intensities of all the maxima remained unchanged and the width of the pattern increased C the intensities of all the maxima decreased and the width of the pattern remained unchanged D the intensities of all the maxima decreased and the width of the pattern increased 4. Monochromatic light of wavelength λ is incident normally on a single slit XY of width a. The diffraction pattern is formed on a screen PQ. The first minimum P makes an angle θ with the direction of the incident light as shown in the figure. X Y θ P Q Monochromatic light Which one of the following gives the correct expression for the path difference (YP – XP), and sin θ? (YP – XP) sin θ A λ 2 λ a B λ 2 2λ a C λ λ 2a D λ λ a 5. A single slit of width d is illuminated by parallel monochromatic light of wavelength λ. A converging lens L of focal length f produces a diffraction pattern on a screen S at the focal plane of the lens as shown in the figure. d L f O x Screen, S The maximum intensity of light on the screen is at O, and the distance of the first minimum from O is x. Which of the following expression for x in terms of d, λ, and f is correct? A x = dλ f C x = df λ B x = λf d D x = d λf 6. A slit of width 0.24 mm is illuminated by parallel monochromatic light. A converging lens of focal length 0.50 m is placed coaxially with the slit, and a diffraction pattern is focussed on a screen. The width of the central maximum on the screen is 3.0 mm. Calculate the wavelength of the monochromatic light. 7. Parallel monochromatic light is incident on a single slit. (a) Sketch a graph of intensity against distance from the centre of the diffraction pattern formed on a screen. (b) Draw a labelled diagram to show the arrangement used to demonstrate diffraction of light by a single slit. Suggest suitable dimensions for the apparatus. 8. The phenomenon of diffraction of light may be demonstrated by illuminating a slit by a parallel beam of monochromatic light. A diffraction pattern is produced on a screen placed at a distance from the slit. (a) Sketch a graph to show the intensity variation in the diffraction pattern. (b) What would happen to the intensity variation if the width of the slit were doubled? 9. The Grand Telescope in the Canary Islands has a diameter of 10.4 m. Assuming that the wavelength of light is 500 nm, what is the resolving power of the telescope? 10. The objective of a microscope has an aperture of diameter 5.0 mm. If an object is view under the microscope using light of wavelength 450 nm, what is the resolving power of the microscope?

Physics Term 3 STPM Chapter 23 Wave Optics 192 23 23.6 Diffraction Gratings Learning Outcomes Students should be able to: • explain the diffraction pattern for a diffraction grating • use the formula d sin θ = mλ for a diffraction grating • describe the use of a diffraction grating to form the spectrum of white light, and to determine the wavelength of monochromatic light 1. A diffracting grating may be obtained by ruling parallel evenly spaced lines of a piece of glass slide. Light passes through the spaces between the ruled lines and produces interference. 2. A diffracting grating may also be an opaque surface with narrow parallel grooves such as that on a compact disc. Light refl ected from the parallel grooves produces interference. 3. The action of a diffraction grating can be explained using Huygens’s principle. A B C Second order m = 2 First order m = 1 Zero order m = 0 First order m = 1 Second order m = 2 Screen Parallel wavefronts Figure 23.28 Diffraction grating 2015/P3/Q17

Physics Term 3 STPM Chapter 23 Wave Optics 193 23 4. Figure 23.28 shows plane light wavefronts incident normally or an idealized diffraction grating consisting of parallel slits. 5. Since the slits are narrow, each slit acts as secondary source of waves. Spherical waves are emitted from the slits in phase. Constructive superposition of waves produces plane wavefronts travelling in the directions shown in Figure 23.29. A bright line is produed on the screen for each of these directions. Monochromatic light Second order line (m = 2) First order line (m = 1) Zeroth order line (m = 0) First order line (m = 1) Second order line (m = 2) Screen θ2 θ2 θ1 θ1 Figure 23.29 6. The central line is known as the zeroth (m = 0) order line. There are the first order line (m = 1), second order line (m = 2), … on each side of the zeroth order line. (Figure 23.29) 7. Figure 23.30 is used to derive an expression for the angle θ1 between the first order line and the zeroth order line. The first order line is produced by constructive interference of waves from neighbouring slits where the path difference between waves from two neighbouring slits such as A and B is equal to λ, the wavelength of the incident light. In Figure 23.30, path difference, BX = λ In triangle AXB, sin θ1 = BX AB = λ d where d = distance between two slits (or lines) on the diffraction grating Hence, d sin θ1 = λ for the first order line. d A B X 2λ Second order θ2 θ2 Figure 23.31 Figure 23.30 d A B X λ First order θ1 θ1

Physics Term 3 STPM Chapter 23 Wave Optics 194 23 8. Figure 23.31 shows the path difference between waves from two neighbouring slit A and B equals to 2λ. Constructive interference between these waves from the slits of the gratings produces the second order line. path difference, BX = 2λ In triangle AXB, sin θ2 = BX AB = 2λ d d sin θ2 = 2λ for second order line (m = 2) 9. In general for the mth order line d sin θm = mλ m = 0, 1, 2, … If N = number of lines per unit length of the diffraction grating, then d = 1 N R V R V R V R V R V m = 2 m = 1 m = 0 m = 1 m = 2 White light Diffraction grating Screen Second order spectrum First order spectrum First order spectrum Zeroth order spectrum White line Second order spectrum = Violet = Red Figure 23.32 10. Figure 23.32 shows a beam of white light incident normally on a diffraction grating. White light consists of various colours, the shortest wavelength is violet and the longest wavelength is red. The zeroth order line is white, but a spectrum is observed for each of the other orders of diffraction. 11. For each order spectrum, the red light is diffracted through a larger angle than violet light.