Question

A laser beam shines along a block of transparent material of length 2.5 m. Part of the beam goes to the detector \( D_1 \) while the other part travels through the block and then hits the detector \( D_2 \). The time delay between the arrivals of the two light beams is inferred to be 6.25 ns. The speed of light \( c = 3 \times 10^8 \, \text{m/s} \). The refractive index of the block is ............ (Round off to two decimal places). 

Hint:

The refractive index can be found by using the time delay and the length of the material in the light path.

Answer 1

Correct Answer: 1.73

Step 1: Relationship between time delay and refractive index.
The time delay \( \Delta t \) between the two light beams can be expressed as the difference in the time it takes for light to travel through the material and the time it takes to travel in air. The time taken to travel through the material is \[ t_{\text{material}} = \frac{L}{v_{\text{material}}}, \] and the time taken in air is \[ t_{\text{air}} = \frac{L}{c}, \] where \( v_{\text{material}} \) is the speed of light in the material, \( L \) is the length of the material, and \( c \) is the speed of light in air. The refractive index \( n \) is related to the speed of light in the material by \[ n = \frac{c}{v_{\text{material}}}. \] Step 2: Calculate the time delay.
The time delay is given as \[ \Delta t = t_{\text{material}} - t_{\text{air}} = \frac{L}{v_{\text{material}}} - \frac{L}{c}. \] Substitute \( v_{\text{material}} = \frac{c}{n} \) into the equation: \[ \Delta t = \frac{L}{\frac{c}{n}} - \frac{L}{c} = L \left( \frac{n - 1}{c} \right). \] Step 3: Solve for the refractive index.
Given that \( L = 2.5 \, \text{m} \) and \( \Delta t = 6.25 \, \text{ns} = 6.25 \times 10^{-9} \, \text{s} \), substitute these values into the equation: \[ 6.25 \times 10^{-9} = \frac{2.5 (n - 1)}{3 \times 10^8}. \] Solve for \( n \): \[ n - 1 = \frac{6.25 \times 10^{-9} \times 3 \times 10^8}{2.5} = 0.75 \Rightarrow n = 1.75. \] Final Answer: The refractive index of the block is \( 1.75 \).