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Solution for Multiple-Choice Homework 23 Optical Systems Solution to Multiple-Choice Homework Problem 23.1(Dropping Coin in Tank) Problem: You are playing a game where you drop a coin into a water tank and try to land it on a target. You often find this game at fast food places as part of a fundraiser. The sides of the tank are flat and the target is 6in from the side of the tank. Your eye is 9in from the side of the tank. Water has index of refraction of 4/3. Because of light refraction at the interface, the target appears, to your eye, to be at a different position than its true position. If you drop the coin accurately and it falls straight down to the location where the target appears to be, how far are you off? Does the coin fall in front or behind the target as you look at it? Select One of the Following: (a) 0.75in behind the target. (b) 1.5in behind the target. (c) 0.75in in front of the target. (d-Answer) 1.5in in front of the target. Solution The object distance is s = 6in. The radius of curvature is infinity because the surface is flat. The incident index is ni = 43 and the transmitted index is nt = 1. The single interface equation is ni nt nt − ni + ′ = r s s Now substitute the quantities given in the problem, 4/3 1 + ′ =0 6in s The image distance is then s′ = −

18 in = −4.5in 4

Therefore, the coin falls 1.5in in front of the target. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 23.2(Two Lens Problem) Problem: A converging lens of focal length 12cm and a diverging lens of focal length −15cm are separated by 55cm. An object is placed 22cm to the left of the converging lens. Compute the distance (with correct sign) the final image forms from the diverging lens. Select One of the Following: (a) +8.1cm (b-Answer) −9.8cm (c) +1.5cm (d) −15cm (e) Toby. Solution (a) Use the Thin Lens Equation for the first lens: 1 1 1 = + ′, f1 s1 s1

1

where f1 = 12cm is the focal length of the first lens, s1 = 22cm is the object distance, and s′ distance of the first lens. Solve for s′1 1 s′1 = 1 . − s1 f s1′ =

1 12cm

is the image

1

1

1 −

1

1 22cm

= 26.4cm

(b) The object distance for the second lens is the separation of the lenses, d, minus the image distance of the first lens, s2 = d − s1′ = 55cm − 26.4cm = 28.6cm. Use the Thin Lens Equation to compute the final image distance, s′2 . 1 , s′2 = 1 1 f2 − s2 where f2 = −15cm is the focal length of the second lens. Solve for s′ 2 : s′2 =

1 f2

1 −

1 s2

=

1 −15cm

1 = −9.8cm 1 − 28.6cm

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 23.3(Shining Light Through a Fish Tank) Problem: You shine a light into a fish tank (the tank is full of water, ignore the effect of the glass) at an angle of θi = 25◦ to the normal, at a point 5cm above the bottom of the tank. Water has an index of refraction of 43 . How far along the bottom of the tank does the refracted ray intersect the bottom? Select One of the Following:

θi θr

5cm

(a) 20cm

d

(b-Answer) 15cm (c) 10cm (d) 5cm (e) Toby Solution The angle of refraction can be found with Snell’s Law

θt = arcsin

µ

ni sin θi = nt sin θt ¶ µ ¶ 1 ni ◦ sin θi = arcsin 4 sin 25 = 18◦ nt 3

The distance along the bottom of the point where light leaves the tank can be found with a little trig. As you can see from the figure to the right tan 18.5◦ = d=

18.5

o

5cm 18.5

o

d

5cm d

5cm = 15cm tan 18.5◦

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 23.4(Radiation Pressure on Mirror at Sun’s Surface) 2

Problem: The power output of the sun is 3.846×1026 W and the radius of the sun (approximately) is 6.96 ×108 m. Calculate the total force at normal incidence on a perfectly reflecting circular mirror of area 1m2 placed at the sun’s surface (before it vaporizes). Select One of the Following: (a-Answer) 0.42N (b) 0.84N (c) 1.7N (d) 7.33N (e) Toby. Solution The intensity of sunlight at the sun’s surface is the power output divided by the surface area of the sun, I=

3.846 × 1026 W P = = 6.32 × 107 W/m2 2 4πr 4π (6.96 × 108 m)2

The radiation pressure on the mirror is I c where the two comes from the perfectly reflecting condition. The force is then Pr = 2

F = Pr A = 2A

6.32 × 107 W/m2 I = 0.42N = 2(1m2 ) c 3 × 108 ms

where A is the area. Still not much. Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 23.5(Focal Length of Diverging Lens) Problem: If an artificial ruby (n = 1.7) was formed into a double convex lens with radii of magnitude 50cm and 75cm, what would the focal length of the lens be? Note, you will have to determine the signs of the radii. Select One of the Following: (a) 0.023cm (b) 12cm (c-Answer) 43cm (d) 210cm (e) Toby. Solution If light reaches the more curved surface first we have r1 = 50cm and r2 = −75cm where the signs are deterined by the location of the radii on the axis.The lens makers equation gives us µ ¶ µ ¶ 1 1 1 1 1 = (n − 1) = (1.7 − 1) = 43cm − − r1 r2 50cm −75cm f

C2

r2

3

C1

0

r1

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 23.6(Light in a Cable) Problem: An Argon laser (λ = 5.0 × 102 nm) shines down a silica glass fiber-optic cable with index of refraction 1.46. What is the wavelength of the laser light, λf , in the cable? Select One of the Following: (a) 220nm (b-Answer) 340nm (c) 500nm (d) 730nm (e) Toby Solution The wavelength of light in a material is reduced from the wavelength in a vacuum by a factor of the index of refraction, 500nm λf = = 340nm. λf = 1.46 n Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 23.7(Two Lens Problem) Problem: In lab, you built a simple telescope out of two convex lenses. Everyone had different lenses, but let’s assume your objective lens had a focal length of f0 = 8cm and your eyepiece fe = 5cm. The lenses are placed the sum of their focal lengths apart. You look through the telescope at an object that is 30cm from the objective lens. Where does the image appear to form in reference to the eyepiece? Report this distance with the correct sign based on our sign convention. Select One of the Following: (a-Answer) −4cm (b) −56cm (c) +11cm (d) +9cm (e) Toby. Solution (a) Use the Thin Lens Equation for the first lens: 1 1 1 + ′, = s1 s1 f1 where f1 = 8cm is the focal length of the first lens, s1 = 30cm is the object distance, and s′ of the first lens. Solve for s′1 1 . s′1 = 1 1 − f1 s1 s′1 =

1 8cm

1

is the image distance

1 = 10.9cm 1 − 30cm

(b) The object distance for the second lens is the separation of the lenses, d, minus the image distance of the first lens, s2 = d − s′1 = 13cm − (10.9cm) = 2.1cm. Use the Thin Lens Equation to compute the final image distance, s′2 . 1 , s′2 = 1 1 − f2 s2 4

where f2 = 5cm is the focal length of the second lens. Solve for s′ 2 : s′2 =

1 f2

1 −

1 s2

=

1 5cm

1 = −3.6cm 1 − 2.1cm

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 23.8(Two Lens Problem) Problem: A concave lens, lens 1, of focal length −50cm and a convex lens, lens 2, of focal length 5cm are separated by 50cm. An object is placed 100cm to the left of lens 1. Compute the magnification of the final image with the correct sign. Light travels through the system from left to right. Light travels from the object passing through lens 1 first and lens 2 second. Select One of the Following: (a) 0.021 (b-Answer) −0.021 (c) 0.043 (d) −0.043 (e) Toby. Solution (a) Image of First Lens: Using the thin lens equation for the first lens 1 1 1 + ′, = s1 s1 f1 where f1 = −50cm is the focal length of the first lens, s1 = 100cm is the object distance for the first lens, and s1′ is the image distance of the first lens. Solve for s′ 1 s′1 = s′1 =

1 −50cm

1 f1

1 −

1 s1

.

1 = −33.33cm 1 − 100cm

The magnification of the first lens is m1 = −

1 −33.33cm s1′ = =− 100cm 3 s1

(b) Object Distance for Second Lens: The object distance for the second lens is the separation of the lenses, d, minus the image distance of the first lens, s2 = d − s′ 1 = 50cm − −33.33cm = 83.33cm. (c) Image of Second Lens: Use the thin lens equation to compute the final image distance, s′ 2 . s′2 =

1 f2

1 −

1 s2

,

where f2 = 5cm is the focal length of the second lens. Solve for s′ 2 : s2′ =

1 f2

1 = − s1 2

1 1 5cm

−

1 83.33cm

= 5.319cm

The magnification of the second lens is m2 = −

5.319cm s2′ = −0.06383 =− 83.33cm s2

The total magnification is the product of the magnification of each element. 1 mT = m1 m2 = ( )(−0.06383) = −0.021 3 5

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 23.9(Wavelength of AM Radio) Problem: You wish to build a very long wavelength radio transmitter. If the target wavelength is 1000m, at what frequency should your transmitter operate? (a) 3.3 × 10−6 Hz (b) 4.8 × 104 Hz (c-Answer) 3.0 × 105 Hz (d) 1.9 × 106 Hz (e) Toby. Solution The wavelength is related to the frequency by c = λf, therefore f=

3 × 108 m c s = = 3.0 × 105 Hz λ 1000m

Total Points for Problem: 3 Points

Solution to Multiple-Choice Homework Problem 23.10(Fish Tank Critical Angle) Problem: You are using a red laser pointer to shine light into a fish tank. What is the maximum angle the light can make with respect to the normal, as the light passes from the glass (n = 1.5) to the water (n = 1.3333)? Select One of the Following: (a) 49◦ (b) 42◦ (c-Answer) 63◦ (d) There is no maximum angle (the maximum angle is 90◦ ). (e) Toby Solution The maximum angle is the critical angle, θc , the angle at which light is refracted at an angle of 90◦ .Use Snell’s law, ni sin θc = nt sin 90◦ = nt . Solve for the critical angle, µ ¶ µ ¶ nt 4/3 −1 −1 θc = sin = sin = 63◦ 3/2 ni Total Points for Problem: 3 Points

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