Question

The refractive index of a transparent material can be determined by measuring the critical angle when the solid is in air. If theta_c = 40.5 degree what is the index of refraction of the material? Tries 0/12 A light ray strikes this material (from air) at an angle of 36.0 degree with respect to the normal of the surface. Calculate the angle of the reflected ray (in degrees). Tries 0/12 Calculate the angle of the refracted ray (in degrees). Tries 0/12 Assume now that the light ray exits the material. It strikes the material-air boundary at an angle of 36.0 degree with respect to the normal. What is the angle of the refracted ray?

Answer 1

1)The index of refraction would be:

n = 1/sin(theta-c)

n = 1/sin40.5 = 1.54

Hence, n = 1.54

2)from Snells law we know that,

n1 sin(theta1) = n2 sin(theta2)

sin(theta2) = n1 x sin(theta1)/n2

sin(theta2) = 1 x sin36/1.54 => theta2 = 22.44 degree

Hence, theta2 = 22.44 degree

4)when the ray exits

n2 sin(theta22.44)/1 = sin(theta3) =>theta3 = 36 deg

Hence, theta3 = 36 deg