Question

In a single slit diffraction pattern, the distance between the plane of the slit and screen is 1.3 m . The width of the slit is 0.65 mm and the second maximum is formed at the distance of 2.6 mm from the centre of the screen. The wavelength of light used is

• A. $6500 \text{ \AA}$
• B. $6000 \text{ \AA}$
• C. $5200 \text{ \AA}$
• D. $4600 \text{ \AA}$

Hint:

Central Maxima width is $2 \lambda D / a$. Secondary Maxima positions are odd multiples of $\lambda D / 2a$.

Answer 1

The Correct Option is C

Step 1: Formula for Secondary Maxima
Position of $n^{th}$ secondary maximum: $y_n = \frac{(2n+1) \lambda D}{2a}$.
For 2nd maximum, $n=2 \implies y_2 = \frac{5 \lambda D}{2a}$.
Step 2: Substitution
$y_2 = 2.6 \times 10^{-3} \text{ m}, D = 1.3 \text{ m}, a = 0.65 \times 10^{-3} \text{ m}$.
$2.6 \times 10^{-3} = \frac{5 \cdot \lambda \cdot 1.3}{2 \cdot 0.65 \times 10^{-3}}$.
Step 3: Calculation
$2.6 \times 10^{-3} = \frac{5 \cdot 1.3 \lambda}{1.3 \times 10^{-3}} \implies 2.6 \times 10^{-3} = 5 \lambda \times 10^3$.
$\lambda = \frac{2.6 \times 10^{-6}}{5} = 0.52 \times 10^{-6} = 5200 \times 10^{-10} \text{ m}$.
Final Answer: (C)