You have three chances to answer this question. Suppose a ray of light enters a

Q: You have three chances to answer this question. Suppose a ray of light enters a

part A : use snells law as na sin A = nb sin B na = 1 nb = 1.89 when incident ray parallel to base A = 60 deg so sin B = 1 * sin 60/1.89 Sin B = 0.458 B = 27.27 deg -------------------------------- here A = ? B = 90-60 = 30 deg A = ? so 1.89 * sin A = sin 30 Sin A = sin 30/1.89 A = 15.34 deg

ID: 1462706 • Letter: Y

Question

You have three chances to answer this question. Suppose a ray of light enters a prism, passes through, and exits on the other side. Assume: - the prism has the shape of an equilateral triangle (60degree at each angle) - the index of refraction of air (surrounding the prism) is 1.00 - the index of refraction of the material of the prism is 1.89 a) If an incident ray travels parallel to the base of the triangle, find the angle of refraction of this ray. HINT: What must the angle of incidence at point A be, in degrees? thetar = degree b) Suppose now the ray hits the surface of the triangle, and now it is the refracted ray that travels parallel to the base of the triangle. In this case, what is the angle of incidence at point A? thetai = degree

Explanation / Answer

part A : use snells law as na sin A = nb sin B

na = 1

nb = 1.89

when incident ray parallel to base A = 60 deg

so sin B = 1 * sin 60/1.89

Sin B = 0.458

B = 27.27 deg

--------------------------------

here A = ?

B = 90-60 = 30 deg

A = ?

so 1.89 * sin A = sin 30

Sin A = sin 30/1.89

A = 15.34 deg

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