Question

As shown in the figure, in Young’s double slit experiment, a thin plate of thickness \(t = 10 \, \mu\text{m}\) and refractive index \(\mu = 1.2\) is inserted in front of slit \(S_1\). The experiment is conducted in air (\(\mu = 1\)) and uses a monochromatic light of wavelength \(\lambda = 500 \, \text{nm}\). Due to the insertion of the plate, central maxima is shifted by a distance of \(x \beta_0\). \(\beta_0\) is the fringe-width before the insertion of the plate. The value of \(x\) is _________.

Hint:

Remember the formula for fringe shift in Young’s double slit experiment when a thin plate is introduced. Ensure consistent units while substituting values.

Answer 1

Correct Answer: 4

To find the shift in the central maxima due to the insertion of the plate, we first calculate the optical path difference introduced by the plate. The formula for optical path difference \(\Delta\) when a plate of thickness \(t\) and refractive index \(\mu\) is inserted is given by:

\[\Delta = (\mu - 1) \times t\]

Substituting the given values, \(t = 10 \, \mu\text{m} = 10 \times 10^{-6} \text{m}\) and \(\mu = 1.2\):

\[\Delta = (1.2 - 1) \times 10 \times 10^{-6} = 0.2 \times 10^{-5} \text{m} = 2 \times 10^{-6} \text{m}\]

The shift in central maxima \(x\beta_0\) is given by:

\[x = \frac{\Delta}{\lambda}\]

Using \(\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \text{m}\):

\[x = \frac{2 \times 10^{-6}}{500 \times 10^{-9}} = \frac{2 \times 10^{-6}}{5 \times 10^{-7}} = 4\]

The value of \(x\) is 4, which lies within the expected range [4, 4].