Question

A slit of width 'a' is illuminated normally by white light for red light ($ \lambda = 6500 \, \text{Å} $). The first minima is obtained at $ \theta = 30^\circ $. Then the value of 'a' is

• A. 3250 \( \text{Å} \)
• B. \( 6.5 \times 10^{-4} \, \text{mm} \)
• C. 1.24 \( \mu \text{m} \)
• D. \( 2.6 \times 10^{-4} \, \text{mm} \)

Hint:

In diffraction problems, remember the formula \( a \sin \theta = m \lambda \) for calculating the slit width.

Answer 1

The Correct Option is C

In the case of single-slit diffraction, the angle of the first minima is given by: \[ a \sin \theta = m \lambda \] For the first minima, \( m = 1 \), and \( \lambda = 6500 \, \text{Å} = 6.5 \times 10^{-7} \, \text{m} \).
Substituting \( \theta = 30^\circ \), we get: \[ a \sin 30^\circ = 6.5 \times 10^{-7} \] \[ a \times \frac{1}{2} = 6.5 \times 10^{-7} \] \[ a = 1.24 \times 10^{-6} \, \text{m} = 1.24 \, \mu \text{m} \]
Thus, the width of the slit is 1.24 \( \mu \text{m} \).