Question

In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle $30^{\circ}$ when light of wavelength $5000 \mathring A $ is incident on the slit. The first secondary maximum is observed at an angle of :

• A. $\sin^{-1} \left( \frac{2}{3} \right)$
• B. $\sin^{-1} \left( \frac{1}{2} \right)$
• C. $\sin^{-1} \left( \frac{3}{4} \right)$
• D. $\sin^{-1} \left( \frac{1}{4} \right)$

Answer 1

The Correct Option is C

To solve this problem, we need to understand the diffraction pattern created by a single slit. The angle for the minima and maxima in a diffraction pattern can be determined using the formulae:

• Minima: a \sin \theta = m \lambda (for minima, m = \pm 1, \pm 2, \ldots)
• Maxima: approximately lies midway between two minima.

Given:

• Wavelength, \lambda = 5000 \mathring A = 5000 \times 10^{-10} \, \text{m}
• The angle for the first minimum, \theta = 30^{\circ}

For the first minimum:

We have a \sin 30^{\circ} = 1 \times 5000 \times 10^{-10}.

Since \sin 30^{\circ} = \frac{1}{2},

a \times \frac{1}{2} = 5000 \times 10^{-10}

Thus, a = 10000 \times 10^{-10} = 10^{-6} \, \text{m}.

The first secondary maximum is approximately at the midpoint between the first and second minima:

The position of the second minimum can be calculated using a \sin \theta_2 = 2 \lambda.

10^{-6} \times \sin \theta_2 = 2 \times 5000 \times 10^{-10}

\sin \theta_2 = \frac{1}{10}

Since this is for the midpoint:

The approximate position for the secondary maximum in terms of angle is where:

\sin \theta_{\text{max}} = \frac{m + \frac{1}{2}} {a/\lambda}, with m = 1.

Calculating \theta_{\text{max}}:

\sin \theta_{\text{max}} = \frac{1.5 \lambda}{a} = \frac{1.5 \times 5000 \times 10^{-10}}{10^{-6}}

= \frac{1.5}{2} = 0.75

Thus, the angle for the first secondary maximum is \sin^{-1} (0.75) = \sin^{-1} \left( \frac{3}{4} \right).

Therefore, the correct option is \sin^{-1} \left( \frac{3}{4} \right).