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Light in vacuum is incident on the surface of a slab of transparent material. In the vacuum the beam makes an angle of 43.8° with the normal to the surface, while in the slab it makes an angle of 19.3° with the normal. What is the index of refraction of the transparent material?

Respuesta :

Answer:

Index of refraction of transparent is 1.48

Step-by-step explanation:

Snell's law states that;

n1(sinθ1) = n2(sinθ2)

Where;

n1 and n2 represent the indices of refraction for the two media, and θ1 and θ2 are the angles of incidence and refraction that the ray R makes with the normal.

In this question;

n1 = 1;

θ1 = 43.8°

θ2 = 19.3°

n2 is unknown.

Thus using Snell's law, we have;

1 x sin 43.8 = n2 x sin 19.3

n2 = (sin 43.8)/sin 19.3

n2 = 1.48

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