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

Step 1: Recall the Formula for Fringe Shift

When a thin plate of thickness \(t\) and refractive index \(\mu\) is introduced in the path of one of the slits in Young’s double slit experiment, the fringe pattern shifts. The fringe shift (\(\Delta x\)) is given by:

\[ \Delta x = \frac{t (\mu - 1)}{\lambda} \beta_0 \]

where \(\lambda\) is the wavelength of light and \(\beta_0\) is the fringe width.

Step 2: Convert Units and Substitute Values

Given \(t = 10 \, \mu \text{m} = 10 \times 10^{-6} \, \text{m}\), \(\mu = 1.2\), and \(\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} = 5 \times 10^{-7} \, \text{m}\):

\[ \Delta x = \frac{10 \times 10^{-6} (1.2 - 1)}{5 \times 10^{-7}} \beta_0 \]

\[ \Delta x = \frac{10 \times 10^{-6} \times 0.2}{5 \times 10^{-7}} \beta_0 \]

\[ \Delta x = \frac{2 \times 10^{-6}}{5 \times 10^{-7}} \beta_0 = \frac{20 \times 10^{-7}}{5 \times 10^{-7}} \beta_0 = 4 \beta_0 \]

Step 3: Find the Value of \(x\)

The central maxima is shifted by a distance of \(x \beta_0\). We found that \(\Delta x = 4 \beta_0\). Therefore, \(x = 4\).

Conclusion: The value of \(x\) is 4.

Answer 2

The correct answer is 4.
Fringe shift $=\frac{t(\mu -1)}{\lambda }B$
$=\frac{10\times 10^{-6}(1.2-1)}{5\times 10^{-7}}B$
$=\frac{10^{-5}\times 0.2}{5\times 10^{-7}}=4$