A ray of light is incident on a pool with three layers of liquid in it at an angle of 75.0°. The top layer is 20.0m deep and has an index of refraction of 1.33. The middle layer is also 20.0m deep and has an index of refraction of 2.00. The bottom layer is 25.0m deep and has an index of refraction of 2.50. At the bottom of the pool there is a mirror that reflects the light back towards the surface. Calculate the time that the ray of light is in the pool, in other words the time taken between points A nd B. Use the speed of light in a vacuum c = 3 * 108 m/s. n₁ = 1.00 n2 = 1.33 n3 = 2.00 75.0⁰ A B 12 = 20.0m 13 = 20.0m

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A ray of light is incident on a pool with three layers of liquid in it at an angle of 75.0°. The top layer is
20.0m deep and has an index of refraction of 1.33. The middle layer is also 20.0m deep and has an
index of refraction of 2.00. The bottom layer is 25.0m deep and has an index of refraction of 2.50. At
the bottom of the pool there is a mirror that reflects the light back towards the surface.
Calculate the time that the ray of light is in the pool, in other words the time taken between points A
and B. Use the speed of light in a vacuum c = 3 * 108 m/s.
Question 2: Waves
n₁ = 1.00
N₂ =
N22
1.33
n3 = 2.00
75.0°
LA
n4 = 2.50
mirror
B
12 = 20.0m
13 = 20.0m
14 = 25.0m
Transcribed Image Text:A ray of light is incident on a pool with three layers of liquid in it at an angle of 75.0°. The top layer is 20.0m deep and has an index of refraction of 1.33. The middle layer is also 20.0m deep and has an index of refraction of 2.00. The bottom layer is 25.0m deep and has an index of refraction of 2.50. At the bottom of the pool there is a mirror that reflects the light back towards the surface. Calculate the time that the ray of light is in the pool, in other words the time taken between points A and B. Use the speed of light in a vacuum c = 3 * 108 m/s. Question 2: Waves n₁ = 1.00 N₂ = N22 1.33 n3 = 2.00 75.0° LA n4 = 2.50 mirror B 12 = 20.0m 13 = 20.0m 14 = 25.0m
Expert Solution
Step 1

To determine: The total time taken between point A and B 

Use the Snell's law at air-water interface 

sinθisinθw=n2n1

where θi is the incident angle, 75o and θw is the angle of refraction at water interface. 

sin75sinθw=1.331.00sinθw=0.726θw=46.55°

From the figure, 

cosθw=l2L2L2=l2cosθwL2=20 mcos46.55L2=29.08 m

The index of refraction is 

n2=cv2v2=cn2

The velocity is ratio of length and time 

L2t2=cn2t2=L2n2ct2=29.08×1.333×108 t2=12.89×10-8 

 

 

 

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