Chapter 23, Problem 6Q

To determine

The speed of light in a solid, rectangular, transparent object

Answer to Problem 6Q

Solution:

The speed of the light in a sold is found by the equation $v=\frac{c}{n_2}$ and measuring the refractive index of the solid from the refraction experiment.

Explanation of Solution

Given:

The width of the solid = y

The angle of incidence = ${\theta }_i$ (Measured from the experiment)

The angle of refraction = ${\theta }_r$ (will be calculated)

The distance from one end of the solid and to the point where the beam enters the solid = x₁

The distance from the same end of solid and to the point the beam exists the solid = x₂

Formula used:

By geometry, the refracted angle can be calculated by the following equation

${\theta }_r={\tan }^{-1}\left(\frac{x_2-x_1}{y}\right)$

The refractive index of the solid $n_2$ can be found by the snell’s law

$\frac{n_2}{n_1}=\frac{\sin {\theta }_i}{\sin {\theta }_r}$

Where,

$n_2\text{is the refractive index of the solid}$

${\text{n}}_1\text{is the refractive index of the air (= 1)}$

If the refractive index of the solid is known, the speed of the light in the solid can be found from the equation.

$n=\frac{c}{v}$ -----3

Where

$\text{c is the speed of light in air or vacuum}$

$\text{v is the speed of light in solid}$

$\text{n is the refractive index of the solid}$

Calculation:

The solid transparent and the rectangular object is the experimental specimen. The light travels into it by beaming the laser beam at a convenient angle ${\theta }_i$ at one side of the solid. The refraction of the light occurs. The following measurements are made.

The angle of incidence is measured.

Measure the distance from one end of the solid to the point where the beam enters the block (x₁). Measure the thickness of the solid (y), and measure the distance from the same end of the solid to the point where the beam exits the solid (x₂).

This is explained by the figure.