As shown in the figure (Figure 1) , a layer of water covers a slab of material X

Question

As shown in the figure (Figure 1) , a layer of water covers a slab of material X in a beaker. A ray of light traveling upwards follows the path indicated. Assume that?1 is a1 = 47.99? and that ?2 is a2 = 67.97?.

Part A

Find the index of refraction of material X . Use 1.38 for the index of refraction of water.

Part B

Find the angle the light makes with the normal in the air.

Part C

What is the critical angle for substance X if the light was leaving it and going out to air?

Part D

What is the speed of light in substance X?

Explanation / Answer

angle of incidence is considered with normal

so angle of incidence = 90 - 67.97 = 22.03 degrees

a) snells law

n1*sin(tehta1) = n2*sin(tehta2)

X*sin(22.03) = 1.33*sin(47.91)

X = 2.63

b) snells law

n1*sin(tehta1) = n2*sin(tehta2)

1.33*sin(47.91) = 1*sin(theta3)

sin(theta3) = 0.987

theta3 = 80.745 degrees

c) snells law

n1*sin(tehta1) = n2*sin(tehta2)

theta2 = 90 as theta1 is critical angle

2.63*sin(tehta1) = 1*sin90

theta1 = 22.3478 degrees

d) n1/n2 = v2/v1

n1 = 1 (air) v1 = 3x10^8 (speed of light in air)

n2 and v2 are refractive index and speed of substance X

n1/n2 = v2/v1

1/2.63 = v2/3x10^8

v2 = 1.14x10^8 m/s

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