1. A ray hits an object at 15° from the vertical. If the material has index of r

Question

1. A ray hits an object at 15° from the vertical. If the material has index of refraction 1.5, what is the refraction angle?
2. If a ray is incident on a block of material at 15° from the vertical, and is refracted to 8° from the vertical, what is the index of refraction?
3. A ray of light crosses from a flint glass slab at 20° from the normal into air. At what angle does it emerge?
4. A ray of light passes through a prism, making the minimum angle of deviation. The rays enter and emerge at 7° from the normal to the surface. What is the index of refraction of the material, if the prism angle is 50°? 

Explanation / Answer

From Snell's law or law of refraction

na sin theta = nb sin theta b

where na is refractive index of medium a and  

nb is refractive index of medium b

theta is angle of incidece and theta b is angle of refraction

1. theta a = 15 degrees , theta b = ?

na = 1 air , nb= 1.5

na *sin theta = nb *sin theta b

1 sin 15 = 1.5 sin theta b

theta b = 9.94 degrees

so the angle of refraction is theta b = 9.94 degrees

2. na = 1 , nb = ? , theta a = 15 degrees , theta b = 8 degrees

susbtituting in the formula  

na *sin theta = nb *sin theta b

1 sin 15 = nb sin theta 8

nb = 1.86

3.

theta a = 20 degrees , na = 1.655 , nb = 1

susbtituting in the formula  

na *sin theta = nb *sin theta b

1.655*sin20 = 1 sin thetab

theta b = 34.47 degrees

4.

A = 50 degrees , Dm = 7 degrees

from the formula of the refractiveindex of prism with angle of prism A and angle of minimum deviation Dm

n = Sin((A+Dm)/2)/ sin (A/2)

n = sin((50+7)/2)/sin(50/2)

n = sin (28.5)/sin25

n = 1.13

so the refractive index of the given prism is n = 1.13

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