PHY 171. Homework 5 solutions. (Due by beginning of class on Wednesday, February 8, 2012)


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1 PHY 171 (Due by beginning of class on Wednesday, February 8, 2012) 1. Consider the figure below which shows four stacked transparent materials. In this figure, light is incident at an angle θ on a boundary between two transparent materials. Some of the light then travels down through the next three layers of transparent materials, while some of it reflects upwards and escapes into the air. Calculate the values of (a) θ 5 and (b) θ 4? (a) Consider the n 1 to n 2 interface in the figure on the right. Since the angle of incidence at this interface is θ 1, the angle of reflection at the n 1 to n 2 interface is also θ 1 (as marked in the figure). From geometry (see figure on the right) the angle of incidence at the n 1 to n 5 interface will then also be θ 1. Note n 5 1, since it is air. Use Snell s law at the n 1 to n 5 interface: n 1 sin θ 1 n 5 sin θ 5 Given that θ , n , and n 5 1 (for air), we get: 1.30 sin sin θ 5 Solving this equation, we find that θ (b) To find θ 4, let us first set up the angles we will be using. Consider again the figure above, where at the n 1 to n 2 interface, we have designated the angle of refraction to be θ 2. Clearly, from geometry (see figure above), the angle of incidence for the n 2 to n 3 interface will then be equal to θ 2. Next, designate the angle of refraction at the n 2 to n 3 interface to be θ 3. Again from geometry, the angle of incidence for the n 3 to n 4 interface will be θ 3. Now, use Snell s law at each interface: At the n 1 to n 2 interface, Snell s law gives n 1 sin θ 1 n 2 sin θ 2, At the n 2 to n 3 interface, Snell s law gives n 2 sin θ 2 n 3 sin θ 3, At the n 3 to n 4 interface, Snell s law gives n 3 sin θ 3 n 4 sin θ 4, We don t need to calculate the intermediate angles θ 2 and θ 3, because things equal to the same thing are equal to one another. That is, we have from the above that so we can write n 1 sin θ 1 n 4 sin θ 4, n 1 sin θ 1 n 2 sin θ 2 n 3 sin θ 3 n 4 sin θ 4 from which we get 1.30 sin sin θ 4, so that θ
2 2. A point source of light is 80.0 cm below the surface of a body of water. Find the diameter of the circle at the surface through which light emerges from the water. If the angle of incidence is equal to the critical angle for the two media (water to air), then the refracted beam will just graze the surface of the water, as shown in the figure on the right. Therefore, light originating from the point source 80.0 cm below the surface that is within the cone whose vertex angle is equal to the critical angle will emerge from the water. On the other hand, light emerging at an angle greater than this will be totally internally reflected back into the water. Using Snell s law, 1.33 sinc 1 sin 90 since the angle of refraction is 90 when the angle of incidence is C, the critical angle. So, ( ) 1 C sin Retain for now more digits than significant to avoid rounding off errors. From geometry, C is also the vertex angle of the cone of light from the point source. This is clear from the figure above, where the twodimensional equivalent of this cone is shown. So with the vertex angle being C, the radius of the circle at the surface from which light emerges is given by r, which can be determined from trigonometry by setting tan C r 80.0 cm from which we obtain r (80.0 cm) tan Therefore, the required diameter of the circle is given by d 2r 2 (80 cm) tan( ) 182 cm 1.82 m Page 2 of 8
3 3. A diverging lens with a focal length of 15.0 cm and a converging lens with a focal length of 12.0 cm have a common central axis. The two lenses are separated by 12.0 cm. An object of height 1.00 cm is 10.0 cm in front of the diverging lens, on the common central axis. (a) Calculate where the lens combination produces the final image of the object. The setup is shown below. Use the lens equations one by one for each lens, keeping the sign convention (given in the Lecture notes) in mind. We will use d id as the image distance for the diverging lens with f d 15 cm. 1 For the diverging lens: 1 1 d o f d d id f d d o f d d o Therefore, d id f d d o d o f d ( 15) ( 15) d id 6 cm Since d id is negative, this means that the image is on the same side of the diverging lens as the object. This image now becomes the object for the converging lens. Since it is on the same side of the diverging lens as the object, we will need to add the distance between the lenses (D 12 cm) to the magnitude of this value (i.e., without considering the sign, since the sign only tells us where the image of the diverging lens alone is located). In other words, the object distance for the converging lens is d oc cm So please be careful, it is obvious that this is one of those problems where you will go wrong if you proceed blindly, without thinking about the signs, etc. Page 3 of 8
4 Moving along, if we use the lens equation again for the converging lens with f c +12 cm (i.e., a positive focal length this time since we are dealing with a convex lens), we get 1 d i 1 f c 1 d oc d oc f c f c d oc Therefore, d i f c d oc d oc f c (12) d i +36 cm This is the position of the final image. Since d i is positive, the image must to the right of the converging lens (i.e., on the opposite side from the diverging lens). (b) What is the height of the image? Show your calculations clearly if you want full credit. Again, proceed one by one. Object height h o 1.0 cm for the diverging lens, so m d d id d o ( 6) ; But m d h id 1.0 cm h id 0.6(1.0) 0.6 cm h id is then also the height of the object h oc for the converging lens, so m c d i (+36) 2; But m c h i d oc cm h i 0.6( 2) 1.2 cm Therefore, the height of the final image is 1.2 cm. The minus sign in the answer above indicates that the image is inverted, as we will discuss in part (d). (c) Is the image real or virtual? Explain. Since the final image formed by the lens combination is to the right of the converging lens (on the opposite side to the object), it must be formed by the light rays actually passing through that location, so it is a real image. (d) Does the image have the same orientation as the object or is it inverted? Explain. In part (b) above, we calculated the magnification for the lens combination, and found it to be negative. This means that the image is inverted. This is as expected the concave lens should keep its image upright, so the object for the convex lens is upright, and the convex lens will produce an inverted image. Page 4 of 8
5 4. In a doubleslit arrangement, the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) What is the angular separation in radians between the central maximum and an adjacent maximum? (a) The condition for a maximum in the 2slit interference pattern is d sinθ mλ where m 0, 1, 2,... is called the order of the maxima, and d is the slit separation. For the central maximum, m 0 in the above equation. For the first offcenter maximum, m 1. The angular separation θ between the central maximum and the first offcenter maximum is then given by d sin θ λ. In this experiment, we are told that d 100λ. So, sin θ λ d which gives θ radians λ 100λ (b) What is the difference between these maxima on a screen 50.0 cm from the slits? The geometry of the problem is shown in the figure below. From the figure, we see that for small θ, we get sin θ tanθ y/l Equating to the value of sinθ in part (a), we get y L so that y L 50.0 cm 0.50 cm Page 5 of 8
6 5. A doubleslit arrangement produces interference fringes for sodium light (λ 589 nm) that are 0.20 apart. What is the angular fringe separation if the entire arrangement is immersed in water (n 1.33)? Once again, the condition for a maximum in the 2slit interference pattern is d sinθ mλ where m 0, 1, 2,... is called the order of the maxima, and d is the slit separation. For the central maximum, m 0 in the above equation. For the first offcenter maximum, m 1. The angular separation θ between the central maximum and the first offcenter maximum is then given by d sin θ λ, which can be rewritten as from which we get d λ/sin d λ sin θ If immersed in water, the frequency remains the same but the speed of light in water changes to a smaller value v water. To find v water, recall that the index of refraction (n) of a medium is related to v water via the relation n c/v water, where c is the speed of light in air. So, putting v water c/n, we get the new wavelength when the arrangement is immersed in water to be But c/f λ, the wavelength in air. λ water v water f Therefore, the wavelength in water is given by c/n f where n 1.33 for water. λ water λ n So, we can write for the new angular fringe separation (θ water ) between the central maximum and the first offcenter maximum that d sin θ water λ water from which we get sin θ water λ water d λ/n sin 0.20 λ/sin 0.20 n sin from which we get the angular fringe separation in water to be θ water 0.15 Page 6 of 8
7 6. The distance between the first and the fifth minima of a singleslit diffraction pattern is 0.35 mm with the screen 40.0 cm away from the slit, when light of wavelength nm is used. (a) Find the slit width. We are given that the distance between the first and fifth diffraction minima is 0.35 mm, the distance from slit to screen is L 40 cm 400 mm, and the wavelength is λ 550 nm mm. Diffraction minima occur at where a is the slit width, and m 1, 2, 3,.... a sin θ m λ (P6.1) As in the interference setup, we have for small θ that sin θ tanθ y/l, where y is the distance from the central point on the screen (i.e., the middle of the central bright fringe) to the position of the minima on the screen. Substituting this relation in equation (P6.1 ), we get ( y a m λ L) from which we get y mlλ/a. So, for the 1 st minimum (m 1), y 1 Lλ a, whereas for the 5th minimum (m 5), y 5 5Lλ a. We are given that y 5 y mm, so So, a 0.35 mm y 5 y 1 5Lλ a Lλ a 4Lλ a 4Lλ 0.35 mm 4 (400 mm) mm 0.35 mm 2.5 mm Therefore, the width of the slit is 2.5 mm (b) Calculate the angle θ of the first diffraction minimum. For the first minimum, a sin θ 1λ. So sin θ λ a mm 2.5 mm This gives θ radians Page 7 of 8
8 7. In a doubleslit experiment, the slit separation d is 2.00 times the slit width w. How many bright interference fringes are there in the central diffraction envelope? We are given that the slit separation, d 2w, where w is the slit width. Let L be the perpendicular distance from the slits to the screen. Since the condition for diffraction minima is w sin θ mλ, where m 1, 2,... (with w here being just a different symbol for the more commonly used a), the first diffraction minima will occur for m 1 at an angle θ D such that sin θ D λ w Therefore, if the first diffraction minimum is at a distance y D from the central spot on the screen, we will get for small θ (for which sin θ D tan θ D y D /L) that ( ) λ y D L w Meanwhile, the condition for interference maxima is d sin θ mλ, where m 0, 1, 2,...; if the m th interference maximum is at a distance y I from the central spot on the screen, we will get for small θ (for which sin θ I tan θ I y I /L) that ( ) ( ) mλ mλ y I L L d 2w To find how many bright interference fringes are in the central diffraction envelope, all we need to do then is set y I y D, and solve for m. This gives ( ) ( ) mλ λ L L 2w w from which we obtain m 2. This means that the 2 nd interference maximum will coincide with the first diffraction minimum. If you look at the figure of how the diffraction envelope sits on top of the interference pattern in the lecture notes, you should realize that this means that we will not see any intensity at this position. So, the only interference fringes we can see within the central diffraction envelope are the central interference maximum, and the first offcenter interference maxima, one to either side of the central maximum. In summary, this means that there are 3 interference maxima visible within the first diffraction envelope. Page 8 of 8
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