Question

In a Young's double slit experiment, the distance between the slits is 1 mm, the slits are at a distance 2 m from the screen, and monochromatic light of wavelength 600 nm is used. If a thin transparent sheet of thickness 0.1 mm and refractive index 1.5 is introduced in front of one of the slits, the interference pattern shifts on the screen by a distance 10 cm.

Answer 1

The statement is correct.

Answer 2

The shift in the central maximum, $\Delta y$, is given by:

$\Delta y = \frac{D}{d}(\mu - 1)t$

Where:

• $D$ is the distance from the slits to the screen
• $d$ is the distance between the slits
• $\mu$ is the refractive index of the sheet
• $t$ is the thickness of the sheet

Given:

• $d = 1 \times 10^{-3} \, \text{m}$
• $D = 2 \, \text{m}$
• $t = 0.1 \times 10^{-3} \, \text{m}$
• $\mu = 1.5$

Substitute the given values into the formula:

$\Delta y = \frac{2}{1 \times 10^{-3}} (1.5 - 1)(0.1 \times 10^{-3}) = 10 \, \text{cm}$

The calculated shift matches the given value.