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 made by the light with normal in medium X = 90 - 67.97 = 22.03 degrees

When a light passes from one medium to another, by snell's law,

sin(i1)/sin(i2) = n2/n1

Applying it here,

sin(22.03)/sin(47.99) = 1.38 / nx

nx = 2.73

(B) If the second medium is air, n of air = 1

So, sin(22.03)/sin(iair) = 1 / 2.73

sin(iair) = 1.024

This is not possible for any i. So, the light reaches critical angle for less than 22.03 degrees in the medium.

(C) Critical angle = sin-1(1/nmedium)

angle = sin-1(1/2.73) = 21.49 degrees.

(D) Speed of light in x = c/nx , where c is speed of light in vacuum

Speed of light in x = 3*10^8 /2.73

= 1.099*108 m/s

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