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

Answer

We know according to optics when A parallel beam of light falls normally on one phase of a small angled glass.

So we can write

$\Delta=\theta(n-1)$ ----(1)

where $\Delta$ is the deviation angle, θ is angle of the prism, n is the refractive index of the glass of given data.

According to condition, we have

$i_1=0$ °

$r_1-i_1=1.8°$

$i_2-r_2=9.2°$

So 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

Using (1), we get as

11.46

(So The refractive index of the glass is equal to 1.96.

And also he angle of the prism is equal to 11.46