Homework-23-sln - OR HW 23 PDF

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OR HW 23...

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Solution for Homework 23 Optical Systems Solution to Homework Problem 23.1(Tunnel in Denver Aquarium) Problem: The Denver Aquarium has a fish tank where the visitor walks through a glassed-in tunnel through the tank. If a visitor looks at a fish at point A, the visitor is looking through a planar interface. If the visitor looks at an eel at point B, then the visitor is looking through a curved interface with radius of curvature 5ft. Both point A and B are 2ft from the tank wall. The index of refraction of water is 4/3.

Water B 2ft

(a)Compute the image location and magnification for a fish at point A. (b)Compute the image location and magnification for the eel at point B . (c)Describe the image of the eel at point B. Justify your description in terms of your calculation.

A

2ft

Elizabeth's฀ Alpha ฀ Fish

Tunnel

Solution to Part (a) For point A, since you are looking through a plane surface the radius of curvature of the surface is ∞. The incident material is water, ni = 4/3. The transmitted material is air, nt = 1. The object, the fish, is a distance s = 2ft from the surface. The single surface refraction equation will locate the image of the fish, nt ni nt − ni + ′. = s r s Substitute, 4/3 1 −1/3 + ′. =0= 2ft s ∞ Solve for the image distance, s′ = −1.5ft. (Quick check, the - is a very good thing, it would be awful if it looked like the fish was out in the air with you...) The magnification for surface A is mA = −

ni s ′ (4/3)(−1.5ft) = 1. =− 2ft nt s

That’s good too, since it is doubtful the aquarium would want the fish to look like they were upside down. It doesn’t look any bigger, but it does look closer, which gives basically the same effect. Grading Key: Part (a) 5 Points Solution to Part (b) For point B, since the center of the curve is on your side (the transmitted side) of the surface, the radius of curvature of the surface is positive. The incident material is water, ni = 4/3. The transmitted material is air, nt = 1. The object, the eel, is a distance s = 2ft from the surface. The single surface refraction equation will locate the image of the eel, ni nt nt − ni = + ′. r1 s s Substitute, 1 4/3 −1/3 + ′. = 2ft 5ft s Solve for the image distance, s′ = −1.36ft. The magnification for surface B is mB = −

ni s ′ (4/3)(−1.36ft) =− = 0.91. nt s 2ft

A little smaller but closer. 1

Grading Key: Part (b) 7 Points Solution to Part (c) The eel looks like he is in the same orientation he really is in (m > 0), this may not be what we would normally consider upright, but hey, it’s an eel. The eel looks slightly reduced, (|m| < 1). The virtual image s′ < 0 means the eel looks like he is still in the water, not in front of the surface with me. Grading Key: Part (c) 3 Points Total Points for Problem: 15 Points

Solution to Homework Problem 23.2(Fish Viewed Through Glass Bottom Boat) Problem: A fish is observed through a thick glass window in the bottom of a boat. One side of the window is flat and one side has radius of curvature r = 3.0ft. The fish is s = 1.0ft below the window. The index of refraction of air is 1, glass 32 , and water 34 . The window is d = 0.50ft thick. Note the diagram is not to scale. This problem may not be worked with the thin lens equation. Where does the fish appear to be to someone on the boat? Describe the image. Be sure to draw any optical coordinate axes along with all important landmarks that you use to solve the problem.

฀ Air

Glass

Water

Solution (a) Draw Optical Axis for each interface: The glass has two surfaces, so we need two axes. surface 2

฀ Air(n=1)

surface 1

r1 r=3ft surface 2

0

Glass(n=3/2) 0 s2’ s2

s1 s1’

surface 1

d=0.5ft s=1ft

Water(n=4/3)

(b) Calculate Image Location and Magnification for Surface 1.: equation. nt1 − ni1 nt1 ni1 + ′ = r1 s1 s1

Use the thick lens (single interface)

The light is incident from water ni1 = 4/3 and is transmitted into the glass nt1 = 3/2 at the first interface. The radius of the curved surface is +3ft. 3/2 − 4/3 1 4/3 3/2 = + ′ = 18ft s1 3ft 1ft

2

27 s′1 = − ft = −1.17ft 23 Calculate the magnification of surface 1 m1 =

−ni1 s1′ −( 34 )(−1.17ft) = nt1 s1 ( 32 )(1ft) = 1.04

(c) Calculate Object Distance for Surface 2: s2 = d − s′1 = 0.5ft − (−1.17ft) = 1.67ft (d) Calculate Image Location and Magnification for Surface 2. : At the second surface light is incident from the glass ni2 = 3/2 and transmitted into air nt2 = 1. The radius of a flat surface is ∞. nt2 nt2 − ni2 ni2 + ′ = =0 s2 s2 r2 µ ¶ 1 nt2 s2 = − 3 (1.67ft) = −1.11ft s2′ = − ni2 2 The magnification of the second interface is m2 = −

ni2 s2′ =1 nt2 s2

flat surface

(e) Compute Total Magnification and Describe Image: mT = m1 m2 = 1.04 Image of the fish is virtual(s′2 < 0), upright(mT > 0), and enlarged(|mT | > 1).. Total Points for Problem: 15 Points

Solution to Homework Problem 23.3(Another Fish Magnifier) Problem: You build a simple device to magnify objects under the water’s surface. The device is a glass rod with curved surfaces on both ends as drawn below. The distance between the vertices of the two surfaces is d = 150cm. The curved surfaces have radii of curvature with magnitude |30cm|. The sign of the each radius of curvature is left to the student. A fish swims 5cm below the surface of the device that is in the water. For this problem use the index of refraction of water, nw = 4/3, and the index of refraction of glass, ng = 3/2. The index of refraction of air is nair = 1 as always. Carry a large number of significant figures through the problem and simplify at the end. (a)Compute the final location of the image formed of the fish by the device. (b)Compute the total magnification of the device.

3

Air

d Glass Water

5cm

Solution to Part (a) This is definitely a “thick” lens, so we must use the single surface refraction equation for each surface. The light is leaving the fish and traveling toward the observer, so at the first surface, the incident material is water, ni1 = 4/3 = nw , and the transmitted material is the glass, nt1 = 3/2 = ng . Since the surface of the glass is concave and light is entering the glass, the radius of curvature is on the incident side, so r1 = −30cm is negative. Computing the location of the intermediate image ”s′ a ” from the first interface, interface 1: ni1 nt1 nt1 − ni1 + ′ = , s1 s1 r1 ng − nw nw ng + ′ = , s1 s1 r1 Substituting in the values given:

3/2 4/3 3/2 − 4/3 + ′ = , 5cm −30cm s1

and s1′ = −5.510cm. Computing the object distance for interface 2, s2 : s2 = d − s2′ = 150cm − (−5.510cm) = 155.51cm. The second interface between the glass and the air also has a center of curvature on the incident side so r2 = −30cm.The incident index for the second surface is the glass ni2 = ng and the transmitted index is the air nt2 = na = 1. Computing the image distance for interface 2, s′ 2 : ni2 nt2 nt2 − ni2 + ′ = s2 s2 r2 and

na − ng ng na + ′ = , s2 s2 r2 1 3/2 1 − (3/2) + , = 155.51cm s2′ −30cm

so s′2 = 142cm. 4

Grading Key: Solution to Part(a) 9 Points Solution to Part (b) The magnification due to surface 1 is m1 = − so m1 = −

ni1 s′1 nt1 s1

(4/3)(−5.510cm) nw s′1 = 0.980. =− (3/2)(5cm) ng s 1

The magnification due to the interface 2 is m2 : ng s′2 ni2 s2′ (3/2)(142cm) =− m2 = − = −1.370. =− na s 2 nt2 s2 (1)(155.51cm) Then the total magnification is found by mT = m1 m2 = (0.980)(−1.370) = −1.3. Grading Key: Solution to Part(b) 6 Points Total Points for Problem: 15 Points

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