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Chapter 35: Q. 20 (page 1017)

A sheet of glass has $n_{r e d} = 1.52$ and $n_{v i o l e t} = 1.55$ . A narrow beam of white light is incident on the glass at $30 掳$ . What is the angular spread of the light inside the glass?

Short Answer

Expert verified

The angular spread of the light inside the glass is $0.39 掳$ .

Step by step solution

01

Given Information.

We have given the refractive index $n_{r e d} = 1.52$ and $n_{v i o l e t} = 1.55$ .

We need to find the angular spread of light inside the glass.

02

Diagram

03

Calculation.

$n_{1} \sin 蟽_{1} = n_{R} \sin 蟽_{R}$

$蟽_{R} = \sin^{- 1} (\frac{n_{1} \sin 蟽_{1}}{n_{R}}) = 19.20$

$n_{1} = 1.00$

$蟽_{1} = 30$

$n_{R} = 1.52$

$蟽_{V} = \sin^{- 1} (\frac{n_{1} \sin 蟽_{1}}{n_{V}}) = 18.82 c w$

$n_{1} = 1.00$

$蟽_{1} = 30$

$n_{V} = 1.55$

$蟽 = 蟽_{R} + 蟽_{V} = 19.20 掳 - 18.82 掳 = 0.385 掳$

Here $n$ is the refractive index and $蟽$ is the refractive angle.

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