l version="1.0" encoding="iso-8859-1"?> Solution Set - Geometric Optics - Physics 106
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Phyllis Fleming Physics

Physics 106

Answers - Geometric Optics

Note that I have used: so = object distance, si = picture distance, Θi for the event ray, Θr for the reflected ray, Θt for the transmitted ray.

You are watching: An angle of refraction is the angle between the refracted ray and the .

1.

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The angle of incidence Θi is the angle in between the incident ray and also the normal N. The angle of reflection Θr is the angle between the reflected ray and the normal N. The angle of refractivity (transmission) Θt is the angle in between the refracted ray and the normal N’.

2.

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When a light ray hits a surchallenge typically, the angle between the event ray and also the normal is 0, so Θi = 0. Since the angle of reflection equals the angle of incidence, Θr= 0. By Snell"s legislation, ni sin Θi = nt sin Θt. For any type of worth of ni and nt, if Θi = 0, Θt = 0.

3.

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The angle of incidence, the angle between the occurrence ray and also the normal N is offered as 45o. Because the angle of reflection equates to the angle of incidence, the angle of reflection is 45o. By Snell"s regulation ni sin Θi = nt sin Θt 1.00 sin 45o = 0.707 = 1.50 sin Θt. sin Θt = 0.707/1.50 = 0.471. Θt = 28o

4.

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Ray 1, designated by one arrow, hits the mirror normally and is reflected ago on itself, as shown in Fig. 4. Ray 2, designated by 2 arrows hits the mirror at some angle of occurrence better than zero and also is reflected ago at the same angle. If you dash the reflected rays behind the mirror, you see that they fulfill at a allude that I have actually labeled I for photo. The big eye checked out at the top of the number sees the two reflected rays as though they have actually originated at the photo. In various other words, if you sight along the two reflected rays they appear to have come from the photo. This image is called a online photo bereason the rays perform not actually satisfy at that allude. The distance of the object O from the mirror, so, is dubbed the object distance and also the distance of the picture I from the mirror, si, is dubbed the image distance. If you meacertain them or usage geometry to calculate the distance si, you uncover that si equates to so for a plane mirror. By geometry, M1OM2 = Θi (different internal angles), OIM2 = Θr (2 parallel lines crossed by a line), and also the angle of incidence = the angle of reflection. Triangles OM1M2 and also IM1M2are congruent (all angles are equal and also they have a common side). Therefore, so = si.

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Using Snell"s legislation to uncover the angle of refraction, 1.00 sin 37o = 1.50 sin Θt. Θt = 23.6o. The angle of transmission and also the angle of incidence on the interchallenge in between the glass and the air are equal (a line cut by two parallel lines). Since Θi = 23.6o at the glass-air interface = Θt at the initial air-glass interconfront, the angle of refraction at the glass-air interchallenge is 37o from Snell"s regulation. I have dashed the incident ray to present you that for a parallel plate, the refracted beam is parallel to the occurrence beam. This tells you that for a very thin parallel plate, the occurrence beam passes through without any deviation, as with a thin lens.

6.

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Using Snell"s law, ni sin Θi = nt sin Qt. At initially interface, 1(sin 50o) = 1.532 sin Θt sin Θt = 0.766/1.532 = 0.500. Θt = 30o. At the second interconfront, as the light ray is going from the glass back into the air, Θ’i = 30o . Now Θ’t = 50o. The path of the ray is shown in Fig. 6 over. For 1 sin 50o = n’ sin Θt, with n’ much less than 1.532, Θt’ would certainly be higher than 30o. The light for λ’ would certainly be bent less towards the normal. When the ray goes from the prism back into the air, n’ sin Θ"i = 1 (sin Θ’t) and also the light ray would be bent much less away from the normal than the light for λ. The all at once impact is that the light of wavesize λ’ has actually a smaller sized deflection from the original direction of the occurrence beam than light of wavesize λ.

7.

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The angle of refractivity Θt = 90o. ni sin Θi = nt sin Θt ni sin 48.6o = 1.00 (sin 90o) ni = 1/0.750 = 4/3 For Θi = 60o, 4/3 sin 60o = (1.00) sin Θt. sin Θt = 4/3 (0.866) = 1.15, which is not feasible bereason the maximum value of a sine is 1. What happens is that the ray is completely internally reflected as displayed in Fig. 7 over via the rays via two arrows.

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From the geomeattempt of the prism, we see that Θ1 = 60o and Θ3 = 30o. (In triangle PBN, tright here are 180o = (30o + 30o + 90o + Θ3). From Snell"s regulation, 1.655 sin Θ3 = 1.333 sin Θ4 (1.655)(0.5000) = 1.333 sin Θ4 Θ4 = 38.4o. Total inner reflection occurs when 1.655 sin Θ1 > nmix sin 90o. Total inner reflection ceases as soon as 1.655 sin 60o = nmix sin 90o = (nmix)(1). nmix = 1.655 (0.866) = 1.433.

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ni sin Θ1 = nt sin Θt. Θi = 90o - 28o = 62o. 1.00 sin 62o = 1.33 sinΘt. sinΘt = 1.00(0.883)/1.33 = 0.664. Θt = 41.6o. tan Θt = 0.888 = 3.0 m/h. h = 3.38 m.

10.

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Ray 1 (indicated with one arrowhead in Fig. for #10) from the scratch at O hits the surface of the slip generally at an angle of incidence of 0o and also continues right into the air with no refraction. Ray 2, (suggested with 2 arrows) is incident upon the surchallenge of the slip at angle Θi and also transmitted at angle Θt. The cam or an eye sees these 2 rays as though they had actually diverged from the photo I. From the geomeattempt of the number, we check out that tan Θt = x/d’ and tan Θi = x/d. From Snell"s legislation, ni sin Θi = nt sin Θt. For sin Θ roughly equal to tan Θ, Snell"s regulation becomes ni tan Θi = nt tan Θt. Or, because tan Θt = x/d’ and tan Θi = x/d, ni x/d = nt x/d’. Thus, ni/nt = d/d’. Putting in nt =1.00, d = 500 µm, and d’ = 500 µm - 120 µm = 380 µm, ni /1 = 500 µm/380 µm = 1.32.

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ni sin Θi = nt sin Θt. For the ray occurrence at the peak of the semicircle (Fig. 9), ni = 1, nt = 4/3, and Θi = 53o. From Snell"s regulation, (1) sin 53o = 4/3 sin Θt or (1)(4/5) = 4/3 sin Θt, sin Θt = 3/5 and Θt = 37o. The transmitted ray travels alengthy a radius, hits the interchallenge in between "water and air" commonly, and experiences no refraction as it goes into the air.

12.

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From the geomeattempt of Fig. 10 above, the angle of incidence at the interconfront in between the prism and air is 45o. For a minimum index of refraction np of the prism, np sin 45o = nair sin 90o or np = 1/sin 45o = 1.41. If the index of refractivity is as well tiny, the refracted ray will be #3 given that it bends amethod from the normal. Ray 1 is the normal and Ray 2 bends towards the normal.

13.

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1/si = 1/f - 1/so = 1/10 cm - 1/30 cm = (3 - 1)/30 cm. si = 15 cm. m = -si/so = -15 cm/30 cm = -1/2. The image is genuine, inverted, and diminiburned in size. 1/si = 1/f - 1/so = 1/10 cm - 1/15 cm = (3-2)/30cm. si = 30 cm. m = -si/so = -30 cm/15 cm = -2. The picture is actual, inverted, and enlarged in dimension. 1/si = 1/f - 1/so = 1/10 cm - 1/5 cm = (1 - 2)/10 cm. si = -10 cm. m = -si/so = +10 cm/5 cm = +2. The image is virtual, erect, and enlarged in size.

14.

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1/si = 1/f -1/so = 1/10 cm -1/30 cm = -(3 + 1)/30 cm. sI = -7.5 cm. m = -si/so = -(-7.5 cm)/20 cm = +0.25. Image is online, erect, diminished. 1/si = 1/f -1/so = 1/10 cm -1/5 cm = -(1 + 2)/10 cm. sI = -10/3 cm. m = -si/so = -(-10/3 cm)/5 cm = +2/3. Image is digital, erect, diminimelted. See ray diagram in number over.

15.

With a solitary converging lens, a digital picture is never before inverted, so the picture in this difficulty need to be actual. The magnitude of the magnification = 3.0 cm/1.0 cm = si/so or si/100 cm = 3 and si = +300 cm since it is a genuine image. 1/so + 1/si = 1/100 cm + 1/300 cm = 4/300 cm = 1/f. f = 75 cm.

16.

6.0 cm/2.0 cm = si/so = si/8.0 cm or the magnitude of si = 24 cm. Due to the fact that it is a virtual photo, we take si = -24 cm. 1/so + 1/si = 1/8.0 cm - 1/24 cm = (3 - 1)/24 cm = 1/12 cm = 1/f. f = 12 cm.

17.

Magnitude of the magnification = 95/5 =19 = si/so or 19so = si (Equation 1) For the projector, the picture is actual. so + si = 4.00 m = 400 cm. (Equation 2) Substituting Eq. 1 right into Eq. 2: so + 19 so = 20 so = 400 cm. so = 20 cm; si = 380 cm. 1/so + 1/si = 1/20 cm + 1/380 cm = 20/380 cm = 1/f. f = 19 cm.

18.

For digital picture, si o + 1/si = 1/12 cm -1/6 cm = -1/12 cm = 1/f. f = -12 cm. f is negative, which synchronizes to a diverging lens.

19.

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Calculate the place of the initially photo. 1/si1 = 1/f1 - 1/so1 = 1/2.0cm - 1/2.2cm. si1 = 22 cm. For minimum eyestrain, the eyepiece is adjusted so that parallel rays emerge, and therefore this picture is at the focal suggest of the eyeitem (Fig. for #19). The total separation of the lenses is 22 cm + 2 cm = 24 cm. If the object"s dimension is y, the genuine photo created at F2 is y’ = y(si1/so1) = y(22 cm/2.2 cm) = 10y. Due to the fact that the eyeitem is changed for minimum eyestrain, its magnifying power is offered by 25/f = 25/2 = 12.5. The over-all magnifying power of the instrument is the product of the two magnifications = M = (10)(12.5) = 125.

20.

First find the 2 focal lengths in meters. fo = 40 cm = 0.40 m 1/fe = 50 D = 50 m-1. fe = 1/50m = 0.02 m. M = fo/fe = 0.40/0.02 = 20. This is a low-power telescope, which is preferable when a big area of the skies is to be scanned.

21.

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1/si = 1/f - 1/so = 1/10cm - 1/30cm = (3 - 1)/30cm. si = 15 cm m = -si/so = -15/30 = -1/2. Image is genuine, inverted, and diminiburned. 1/si = 1/f - 1/so = 1/10cm - 1/5cm = (1 - 2)/10cm. si = -10 cm. m = -si/so = -(-10cm)/5cm= + 2. Image is digital, erect, and also enlarged. In the ray diagrams for the concave mirror, (1) a ray attracted parallel to the primary axis is reflected ago with the focal point, (2) a ray attracted through the facility of curvature hits the mirror commonly (angle of incidence = 0) and also is reflected earlier on itself (angle of reflection = 0), and (3) a ray attracted through the focal allude is reflected earlier parallel to the primary axis.

22.

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In the figure above, parallel rays event on the converging lens pass through its focal point F’1 and also the focal point F2 of the concave mirror, hit the mirror, and also are reflected back parallel to the primary axis. After returning through the converging lens they pass with its focal suggest at F1.

23.

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From the geomeattempt of Fig. 11 above, L1 + L2 = (a2 + x2)1/2 + 1/2 t = L1/v1 + L2/v2 t = (a2 + x2)1/2/(c/n1) + 1/2/(c/n2) dt/dx = n1x/c(a2 + x2)1/2 - n2(d - x)/c1/2 For a minimum, dt/dx = 0 or n1x/(a2 + x2)1/2 = n2(d - x)/1/2 n1 sin Θ1 = n2 sin Θ2

24.

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For the initially photo, 1/si1 = 1/f1 - 1/so1 = 1/10cm - 1/15cm = (3 - 2)/30cm = 1/30cm. si1 = 30 cm. With a distance of 30 cm from the first lens and a lens separation of 20 cm, the picture of the first lens (had actually the second lens not been there) would be created 10 cm to the right of the second lens. We say this object distance for the second lens is virtual and also designate this by taking so2=-10cm. Then 1/si2 = 1/f2 - 1/so2 = 1/10cm - (-1/10cm) = 2/10 cm, or si2=5cm to the right of the second lens. For the initially lens, m1 = -si1/so1 =-30/15 = -2. For the second lens, m2 = -si2/so2 = -5/(-10) = +1/2. Total magnification = m1m2 =(-2)(1/2) = -1. In Fig. for #24 over, ray 2 is drawn from the peak of the object at O’ through the facility of the lens. Ray 3 is drawn with the focal allude of the initially lens F1 and is refracted through the first lens parallel to the primary axis. If the second lens had not been tbelow, these 2 rays would accomplish at the top of the initially photo I’1. Any various other ray that starts from O’ will likewise be refracted with the initially lens to arrive at I’1. Ray 4 in Fig. for #24 that passes through the center of the second lens will continue tright here also via the second lens. (You draw it backwards. Starting at I’1 attract a line that passes via the center of the second lens until it hits the axis of the first lens. Then attract a line from O’ to the suggest wright here it hits the axis of the first lens.) Ray 3 that arrives at the second lens parallel to the major axis will be refracted with the focal suggest of the second lens. Rays 3 and also 4 accomplish at the height of the second photo at I’2. Therefore the second image appears as displayed in Fig. for #24 as I2I’2. The final photo is actual (note si2 > 0), inverted (initially lens inverted the image, but second lens did not invert that image), so the last image is the inverted and also the same dimension as the original object (m1m2) = -1.

25.

1/fred = (n – 1)(1/R1 + 1/R2) = (1.51 – 1)(1/38 cm + 1/38 cm) = 0.0268 cm-1 1/f – 1/so = 0.0268 cm-1 – 1/95 cm = 1/si = 0.0163 cm-1. si = 61.3 cm. 1/fblue = (n – 1)(1/R1 + 1/R2) = (1.54 – 1)(1/38 cm + 1/38 cm) = 0.0284 cm-1 1/f – 1/so = 0.0284 cm-1 – 1/95 cm = 1/si = 0.0179 cm-1. si = 55.9 cm. You can buy a electronic camera via an acromatic lens that corrects for the variation in the index of refractivity through shade. This problem shows the expense may be worth it.

26.

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Due to the fact that we want the farthest allude for the eye to be at infinity, we take so equal to ∞. Since we require these rays to show up to come from 4.0 m, the far suggest, we take si to be -4.0 m. The picture distance is negative because the rays perform not really fulfill there, however show up to the eye to come from there. 1/so + 1/si = 1/∞ + -1/4.0m = 1/f. f = -4.0m. Diopters = 1/f = -0.25 D.

27.

Now we desire the rays to appear to come from 1.25 m once we location an object at the welcomed close to point of 0.25 m. so = 0.25 m and also the digital photo position si = -1.25m. 1/so + 1/si = 1/0.25m + -1/1.25m = 1/f. f = +0.312 m. Diopters = +3.2 m-1 = + 3.2 D.
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