Question

In Young's double slit experiment, when a screen is placed at a distance of 1.2 m from the plane of the slits, the fringe width is 2 mm. If the setup is immersed in a liquid of refractive index 1.25 and the screen is moved to a distance of 1.8 m from the slits, then the fringe width is

• A. 2.4 mm
• B. 1.6 mm
• C. 2.1 mm
• D. 1.8 mm

Hint:

Fringe width increases with screen distance and decreases with refractive index of the medium.

Answer 1

The Correct Option is A

Concept: Fringe width in Young's double slit experiment is \[ \beta=\frac{\lambda D}{d} \] When the entire arrangement is immersed in a medium of refractive index \(\mu\), \[ \lambda'=\frac{\lambda}{\mu} \] Therefore, \[ \beta'= \frac{\beta}{\mu} \cdot \frac{D'}{D} \]

Step 1: Write the given data.
\[ \beta=2\text{ mm} \] \[ D=1.2\text{ m} \] \[ D'=1.8\text{ m} \] \[ \mu=1.25 \]

Step 2: Calculate the new fringe width.
\[ \beta' = 2 \times \frac{1.8}{1.2} \times \frac{1}{1.25} \] \[ = 2\times1.5\times0.8 \] \[ = 2.4\text{ mm} \] \[ \boxed{2.4\text{ mm}} \]