Answer to Question #97086 in Optics for Duah Gideon

Question

A parallel beam of light falls normally on one phase of a small angled glass
prism. Part of the beam is refracted at the second phase and is deviated
through an angle of 1.8o. Another part is reflected at the second phase and
emerges at the first making an angle of 9.2o with the normal. Calculate:
(i) The refractive index of the glass.
(ii) The angle of the prism.

Answer 1

A parallel beam of light falls normally on one phase of a small angled glass.

In this case, we can write

$δ=θ(n-1) (1)$

where δ is the deviation angle, θ is angle of the prism, n is the refractive index of the glass

 

According to condition, we have

i₁=0°

r₁-i₁=1.8°

i₂-r₂=9.2°

According to Figure, we have

r₂=r₁=1.8°

The deviation angle is equal to

δ=9.2+1.8=11°

According to Snellus's law, we get

$\frac {\sin {i_2}}{\sin {r_2}}=n (2)$

Using (1) we get

n=1.96

According to (1), we get

11.46

Answer

(i) The refractive index of the glass is equal to 1.96

(ii) The angle of the prism is equal to 11.46