Question

A laser beam shines along the surface of a block of transparent material. (See the figure .) Half of the beam goes straight to a detector, while the other half travels through the block and then hits the detector. The time delay between the arrival of the two light beams at the detector is 6.25 ns. And has a length of 2.50 m

Answer 1

Concepts and reason

The concept required to solve the given problem is velocity of the light in the medium and time taken by the light to travel a distance.

Initially, derive the expression for calculating time taken by light to travel distance (d) in air and in the block and finally index of refraction of the material will be calculated.

Fundamentals

Index of refraction is defined as the ratio of the speed of light in medium 1 and the speed of light in medium 2.

The expression for the index of refraction is,

$n = \frac{{{v_1}}}{{{v_2}}}$

Here, n is the index of refraction of medium 2 with respect to medium 1, v₁ is the speed of light in medium 1 and v₂ is the speed of light in medium 2.

The velocity of light in the block is,

${c_b} = \frac{c}{n}$

Here, ${c_b}$ is the velocity of light in block, c is the speed of light in vacuum, and n is the refractive index of the material of the block.

The time taken by light to travel a distance d is,

$t = \frac{d}{v}$

Here, d is the distance traveled by the object and v is the velocity of the object.

Time taken by light to light to travel a distance (d) in air is,

${t_a} = \frac{d}{c}$ …… (1)

Here, t_a is the time taken by light to travel a distance d in air, d is the distance covered by light, and c is the speed of light in air.

Time taken by light to travel a distance (d) in the block is,

${t_b} = \frac{d}{{{c_b}}}$

Here, d is the distance covered by light, c_b is the velocity of light in block and t_b is the time taken by light to travel a distance (d) in the block.

Substitute $\frac{c}{n}$ for ${c_b}$.

${t_b} = \frac{{nd}}{c}$ …… (2)

From equations (1) and (2), time delay between arrivals of the two light beams at the detector is,

$\begin{array}{c}\\{t_b} - {t_a} = \frac{{nd}}{c} - \frac{d}{c}\\\\ = \left( {n - 1} \right)\frac{d}{c}\\\end{array}$

Rearrange the above equation as follows:

$\left( {n - 1} \right) = \frac{{c\left( {{t_b} - {t_a}} \right)}}{d}$

Substitute $6.25 \times {10^{ - 9}}{\rm{ s}}$ for $\left( {{t_b} - {t_a}} \right)$, 2.50 m for d, and $3 \times {10^8}{\rm{ m/s}}$ for c.

$\begin{array}{c}\\\left( {n - 1} \right) = \frac{{\left( {3 \times {{10}^8}{\rm{ m/s}}} \right)\left( {6.25 \times {{10}^{ - 9}}{\rm{ s}}} \right)}}{{2.50{\rm{ m}}}}\\\\ = 0.75\\\\n = 1.75\\\end{array}$

Ans:

The index of refraction of material is 1.75.