1. The refractive index of a transparent material can be determined by measuring

Question

1. The refractive index of a transparent material can be determined by measuring the critical angle when the solid is in air. If ?c= 41.5° what is the index of refraction of the material?

2. A light ray strikes this material (from air) at an angle of 36.9° with respect to the normal of the surface. Calculate the angle of the reflected ray (in degrees).

3. Calculate the angle of the refracted ray (in degrees).

4. Assume now that the light ray exits the material. It strikes the material-air boundary at an angle of 36.9° with respect to the normal. What is the angle of the refracted ray?

Explanation / Answer

Ans 1:

We know that according to snells law

n1sin?1 = n2sin?2 where n is refractive index and ? is angle

at critical angle we have ?2 = 90 deg

1*sin(41.5) = n2

n2 = 0.66

Ans 2 :

Angle of reflected rai = Angle of incidient ray = 36.9

Ans 3:

Applying snell's law we have:

1*sin(36.9 ) = 0.66*sin(?)

sin? = sin(36.9)/0.66 = 0.6/0.66 = 0.909

? = arcsin(0.909) = 65.37 ged

Ans 4

Applying snells law we have

0.66sin(36.9) = 1*sin?

sin? = 0.396

? = 23.33

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