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The refractive index of a transparent material can

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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

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