Question

In Young's double slit experiment, fringe width is \( 1.4 \text{ mm} \) with light of wavelength \( 6000\,\text{\AA} \). If the light of wavelength \( 5400\,\text{\AA} \) is used, with no other change in the experimental set up. The change in fringe width is

• A. $1.26 \text{ mm}$
• B. $0.12 \text{ mm}$
• C. $0.13 \text{ mm}$
• D. $0.14 \text{ mm}$

Hint:

- $\beta \propto \lambda$ - Decrease in wavelength $\Rightarrow$ decrease in fringe width

Answer 1

The Correct Option is D

Concept: Fringe width in YDSE: \[ \beta = \frac{\lambda D}{d} \Rightarrow \beta \propto \lambda \]

Step 1: Use proportionality.
\[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} = \frac{5400}{6000} = 0.9 \]

Step 2: Find new fringe width.
\[ \beta_2 = 1.4 \times 0.9 = 1.26\ \text{mm} \]

Step 3: Calculate change.
\[ \Delta \beta = 1.4 - 1.26 = 0.14\ \text{mm} \] Answer: $0.14 \text{ mm}$