Question

In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is $5\ \text{mm}$. The screen on which the diffraction pattern is displayed is at a distance of $80\ \text{cm}$ from the slit. The wavelength is $6000\ \text{\AA}$. The slit width is about

• A. $0.576\ \text{mm}$
• B. $0.348\ \text{mm}$
• C. $0.192\ \text{mm}$
• D. $0.096\ \text{mm}$

Hint:

The distance between the first minimum on the left and the first minimum on the right is simply the full angular width of the central maximum projected onto the screen ($2\lambda D / a$). Don't confuse it with the standard fringe width of double-slit interference patterns!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the distance between the first minima on either side of the central maximum in a single slit diffraction pattern. We need to calculate the width of the slit.

Step 2: Key Formula or Approach:
The position of the $n$-th minimum on the screen from the central maximum is $y_n = \frac{n\lambda D}{a}$.
The distance between the first minimum on the left and the right is the total width of the central maximum, which is $2y_1 = \frac{2\lambda D}{a}$.
Rearranging this to solve for slit width gives $a = \frac{2\lambda D}{2y_1}$.

Step 3: Detailed Explanation:
Given parameters:
Distance between first minima, $2y_1 = 5\ \text{mm} = 5 \times 10^{-3}\ \text{m}$.
Distance to screen, $D = 80\ \text{cm} = 0.8\ \text{m}$.
Wavelength, $\lambda = 6000\ \text{\AA} = 6000 \times 10^{-10}\ \text{m} = 6 \times 10^{-7}\ \text{m}$.
Substitute these values into the formula:
$$5 \times 10^{-3} = \frac{2 \times 6 \times 10^{-7} \times 0.8}{a}$$
$$a = \frac{9.6 \times 10^{-7}}{5 \times 10^{-3}}$$
$$a = 1.92 \times 10^{-4}\ \text{m}$$
Convert the slit width into millimeters to match the options:
$$a = 0.192 \times 10^{-3}\ \text{m} = 0.192\ \text{mm}$$

Step 4: Final Answer:
The slit width is about $0.192\ \text{mm}$, matching option (C).