Question

In Young's double slit experiment, the distance between screen and aperture is $1\text{ m}$ . The slit width is $2\text{ mm}$ . Light of $6000\text{ \AA}$ is used. If a thin glass plate ( $\mu = 1.5$ ) of thickness $0.04\text{ mm}$ is placed over one of the slits, then there will be a lateral displacement of the fringes by}

• A. 0.5 cm
• B. 1 cm
• C. 1.5 cm
• D. 2 cm

Hint:

Lateral shift $\Delta x = \beta \frac{(\mu - 1)t}{\lambda}$, where $\beta$ is fringe width. The pattern always shifts toward the side where the glass plate is placed.

Answer 1

The Correct Option is B

Step 1: Concept
When a transparent plate is introduced in the path of one wave, the fringe pattern shifts by a distance $\Delta x = \frac{D}{d}(\mu - 1)t$.

Step 2: Meaning
$D = 1 \text{ m}$, $d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$, $\mu = 1.5$, $t = 0.04 \text{ mm} = 4 \times 10^{-5} \text{ m}$.

Step 3: Analysis
$\Delta x = \frac{1}{2 \times 10^{-3}} \times (1.5 - 1) \times (4 \times 10^{-5})$
$\Delta x = \frac{1}{2 \times 10^{-3}} \times 0.5 \times 4 \times 10^{-5} = \frac{2 \times 10^{-5}}{2 \times 10^{-3}}$
$\Delta x = 10^{-2} \text{ m} = 1 \text{ cm}$.

Step 4: Conclusion
The lateral displacement of the fringes is $1 \text{ cm}$. Final Answer: (B)