2t = (m+ 1 /2) λ = (m+ 1 /2)(λ/n); min, m = 0, 1, 2,... n1 < n2 < n3 2t = m λ = m(λ/n); min, m = 0, 1, 2,... n1 < n2 > n3

Transcription

1 PHY1160C Exam #3 July 8, 1997 Possibly useful information: For reflection, θinc = θref For refraction, image equation apparent depth Young s Double Slit: n1 sin θ1 = n2 sin θ2 n = c/v M = h i = d i h o d o 1 f = d i d o di = do (n1/n2) = d sin θ d sin θ = m λ, maxima for m = 0, 1, 2,... d sin θ = (m + 1 /2) λ, minima for m = 0, 1, 2,... Diffraction Grating: = d sin θ d sin θ = m λ, maxima for m = 0, 1, 2,... Diffraction by a slit: = D sin θ First minimum for = D sin θ = λ sin θ = λ/d θ λ/d Minima for = D sin θ = m λ, m = 1, 2, 3,... Optical Resolution: θ 1.22( λ/d ) Thin Films: λ = λ/n 2t = m λ = m(λ/n); max, m = 0, 1, 2,... n1 < n2 < n3 Simple Magnifier: 2t = (m+ 1 /2) λ = (m+ 1 /2)(λ/n); min, m = 0, 1, 2,... n1 < n2 < n3 2t = m λ = m(λ/n); min, m = 0, 1, 2,... n1 < n2 > n3 2t = (m+ 1 /2) λ = (m+ 1 /2)(λ/n); max, m = 0, 1, 2,... n1 < n2 > n3 Mang = r / f = 25 cm / f Microscope: M tot = L f eye f obj 25 cm f eye Telescope: M ang = θ θ = f obj f eye Key #3, p - 1 -

2 I. Consider an object placed 30 cm in front of a converging lens whose focal length is f = 10 cm. Using the image equation or better still drawing a ray diagram in the space provided, describe the image. F F di > 0 which means this is a real image. M = di /do = 15 cm /30cm = 0.5 M < 0 means the image is inverted. M < 1 means the image is smaller. 1. The image is A) virtual, inverted B) real, inverted C) virtual, upright D) real, upright 1 / f = 1 /di + 1 /do 1 / di = 1 /f 1 /do 1 / di = 1 /10 cm 1 /30 cm 1 / di = ( ) / cm 1 / di = / cm di = ( 1 /0.067) cm di = 15 cm 2. The (absolute value of the) magnification is A) M = 0.5 B) M = 1.0 C) M = 2.0 D) M = The image is located at A) di = 10 cm B) di = 15 cm C) di = 20 cm D) di = 30 cm Key #3, p - 2 -

3 II. Consider an object placed 30 cm in front of a convex mirror whose focal length is f = 10 cm. Using the image equation or better still drawing a ray diagram in the space provided, describe the image. F 1 / f = 1 /di + 1 /do 1 / di = 1 /f 1 /do 1 / di = 1 /( 10 cm) 1 /30 cm 1 / di = ( ) / cm 1 / di = / cm di = ( 1 /0.133) cm di = 7.5 cm di < 0 which means this is a virtual image. M = di /do = ( 7.5 cm) /30cm = M > 0 means the image is upright. M < 1 means the image is smaller. 4. The image is A) virtual, inverted B) real, inverted C) virtual, upright D) real, upright 5. The (absolute value of the) magnification is A) M = 0.25 B) M = 0.50 C) M = 2.0 D) M = The image is located at A) di = 7.5 cm B) di = 10 cm C) di = 15 cm D) di = 30 cm Key #3, p - 3 -

4 III. A Helium-Neon (HeNe) laser produces red light with a wavelength of 634 nm (that is, 634 x 10 9 m). Consider such a HeNe laser that shines light onto a diffraction grating with 6500 lines per centimeter. 7. What is the spacing between the lines on the diffraction grating? 8. At what angle θ will the beam be diffracted? 9. On a screen 1.0 m from the grating, where will you see the diffracted beam? That is, at what linear distance from the zeroth order or undiffracted beam, will you find the first order diffracted beam? θ =? θ y =? 1.00 m d = ( 1 cm /6500) d = x 10 4 cm d = x 10 4 cm [ 1 m / 100 cm ] d = x 10 6 m There will be diffraction maxima for d sin θ = m λ We are interested in the first order maximum; that is for m = 1 d sin θ = λ sin θ = λ / d sin θ = (634 x 10 9 m) / (1.538 x 10 6 m) sin θ = θ = 24.3 tan θ = opp / adj = y / 1.0 m y = (1.0 m) tan θ y = (1.0 m) tan 24.3 y = (1.0 m) (0.452) y = 0.45 m Key #3, p - 4 -

5 7. What is the spacing between the lines on the diffraction grating? A) d = x 10 9 m B) d = x 10 6 m C) d = x 10 3 m D) d = x 10 6 m 8. At what angle θ will the beam be diffracted? A) θ = 13.2 B) θ = 24.3 C) θ = 41.2 D) θ = On a screen 1.0 m from the grating, where will you see the diffracted beam? A) 0.25 m B) 0.35 m C) 0.45 m D) 0.55 m Key #3, p - 5 -

6 10. A ray of light strikes a mirror with an angle of incidence of 53. It is reflected with an angle of reflection that equals θi θr The Law of Reflection is that θi = θr so the angle of reflection θr is also 53. A) 37 B) 45 C) 53 D) A ray of light strikes a piece of glass (n = 1.33) with an angle of incidence of 37. It is refracted with an angle of refraction that equals n1 = nair = 1.00 θi n2 = nwater = 1.33 θr Snell s Law describes refraction at a plane interface n1 sin θ1 = n2 sin θ2 A) 14 B) 27 C) 35 D) 53 (1.00)(sin 37 ) = (1.33)(sin θ2) (1.00)(0.60) = (1.33)(sin θ2) sin θ2 = 0.60 / 1.33 sin θ2 = θ2 = 26.8 θ2 27 Key #3, p - 6 -

7 12. A fish in an acquarium (n = 1.33) looks like it is 20 cm below the water s surface. How far, actually, is the fish below the water s surface? (You are viewing the fish from air, of course). The fish s apparent depth is 20 cm. The apparent depth di and actual depth do are related by di = do [ n air / nwater ] A) 15 cm B) 20 cm C) 27 cm D) 54 cm di = do [ 1.0 / 1.33 ] di = 0.75 do do = 1.33 di do = (1.33) (20 cm) do = 26.6 cm do 27 cm 13. The image of a real object produced by a concave mirror... A) is always real. B) may be real or virtual. C) is always virtual and inverted. D) is always virtual. 14. The image of a real object produced by a convex mirror... A) is always real. B) may be real or virtual. C) is always virtual and inverted. D) is always virtual. 15. The image of a real object produced by a converging lens... A) is always real. B) may be real or virtual. C) is always virtual and inverted. D) is always virtual. 16. The image of a real object produced by a diverging lens... A) is always real. B) may be real or virtual. C) is always virtual and inverted. D) is always virtual. Key #3, p - 7 -

8 17. A real image formed by a single lens is always A) enlarged and inverted. B) reduced in size and upright. C) inverted but may be enlarged or reduced. D) reduced in size but may be inverted or upright. 18. A virtual image formed by a single lens is always A) enlarged and upright. B) reduced in size and inverted. C) upright but may be enlarged or reduced. D) enlarged but may be inverted or upright. 19. Two ordinary light bulbs show an interference pattern like Young s double slit... A) if they are close enough. B) if they are each behind a card with a slit in it. C) if they are behind two slits in a single card. D) if they have long, straight filaments. E) none of the above (they are incoherent sources) 20. A microscope is made from an objective lens with a focal length of 0.50 cm and an eyepiece with a focal length of 0.75 cm. The tube length or distance between the lenses is 18.0 cm. What is the magnification of the microscope? A microscope s magnification is given by M tot = L f eye f obj 25 cm f eye Mtot = [ (18.0 cm 0.75 cm) / 0.50 cm ] [ 25 cm / 0.75 cm ] Mtot = [ / 0.50] [ 33.3 ] Mtot = [ 34.5] [ 33.3 ] Mtot = 1149 Mtot 1150 A) 36 x B) 575 x C) M = 1150 x D) 2300 x Key #3, p - 8 -

9 21. What is the magnification of a telescope made with an objective lens with a focal length of 45 cm and an eyepiece with a focal length of 1.5 cm? A telescope s magnification is given by M ang = θ θ = f obj f eye M = 45 cm / 1.5 cm M = 30 A) 15 x B) M = 30 x C) 45 x D) 67.5 x 22. Light from a ruby laser (λ = nm) passes through a double slit with separation of 0.10 mm and then forms an interference pattern on a screen 2.5 meters away. How far or at what angle from the central maximum or bright spot is the next maximum or bright spot? We know the following: L = 2.5 m d = 0.10 mm = 1.0 x 10 4 m λ = nm = x 10 7 m We expect to find the first maximum at angle θ given by d sin θ = m λ with m = 1. d sin θ = m λ = λ d sin θ = λ sin θ = λ /d θ = 0.40 Realistically, however, it would be easier to measure the linear position of this first maximum. It is straightforward to use this to find the linear position, the y- value. The tangent of the angle θ is given by tan θ = y/l y = L tan θ y = (2.5 m) tan(0.398 ) y = (2.5 m)(6.943 x 10 7 ) y = 1.7 x 10 2 m y = 1.7 cm Notice that for such small angles the sine and the tangent are indistinguishable. The first maximum is located 1.7 cm from the central maximum. A) 0.40 B) 4.0 C) 14.0 D) 40.0 Key #3, p - 9 -

10 23. A white light source like a long, straight filament of a showcase bulb is viewed through a diffraction grating whose grating spacing is d = 1.0 x 10 5 m (that is, the grating has 1,000 lines per centimeter). A colorful spectrum is seen. The angular separation between the red (λ = 700 nm) and blue (λ = 400 nm) parts of the spectrum is... We know the following: d = 1.0 x 10 5 m λ(red) = 700 nm = 7 x 10 7 m λ (blue) = 400 nm = 4 x 10 7 m For each color, we expect to find a bright area, or the first maximum, at angle θ given by d sin θ = m λ with m = 1. d sin θ = m λ = λ d sin θ = λ sin θ = λ /d For red, with λ(red) = 700 nm = 7 x 10 7 m, this is sin θred = 7 x 10 7 m / 1.0 x 10 5 m = 7 x 10 2 = 0.07 θred = 4.01 For blue, with λ(blue) = 400 nm = 4 x 10 7 m, this is sin θblue = 4 x 10 7 m / 1.0 x 10 5 m = 4 x 10 2 = 0.04 θblue = 2.29 The difference in these angles is θ = = 1.72 A) θ = = 16.1 B) θ = = 21.4 C) θ = = 8.1 D) θ = = 1.72 Key #3, p

11 24. Yellow light, with λ = 560 nm, shines on a soap bubble. When viewed under near normal incidence, the soap bubble appears very bright. The index of refraction of the soap and water solution is n = 1.4. What is the thickness of the soap bubble at this bright area? With air film air, there is a phase shift so we find a maximum or a bright area when the total distance traveled, 2t, is one half a wavelength (one-half of the wavelength inside the soap film), n = 1.0 t n = 1.4 n = 1.0 A) 0.62 x 10 7 m B) 1.23 x 10 7 m C) 1.73 x 10 7 m D) 2.46 x 10 7 m E) none of the above 2t = ( 1 /2) λ 2t = ( 1 /2) (λ/n) t = ( 1 /4) (λ/n) t = ( 1 /4) (560 nm/1.4) t = 100 nm t = 1.00 x 10 7 m The original solution was in error for this question so everyone s score has been increased by 4 points! Key #3, p

12 25. What sort of corrective, eyeglasses lens should be prescribed for a nearsighted person? A) converging B) diverging Key #3, p