Problems on Diffraction Grating – Solved Numericals

1. Find the largest wavelength of light falling on double slits separated by 1.20 µm for which there is a first-order maximum.

Given:

• Distance between slits, \( d = 1.2 \times 10^{-6} \, \text{m} \)
• Order of diffraction, \( n = 1 \)

The condition for maxima is

\[ n\lambda = d \sin\theta \]

The largest wavelength occurs when \( \sin\theta = 1 \).

\[ \lambda = d = 1.2 \times 10^{-6} \, \text{m} \] \[ \lambda = 1200 \, \text{nm} \]

This wavelength lies in the infrared region, not in the visible spectrum.

2. A grating containing 4000 slits per centimetre produces a second-order bright line at an angle of 30°. Find the wavelength of light used.

Given:

• Number of slits = 4000 per cm
• Grating element, \[ d = \frac{1}{4000} \, \text{cm} = 2.5 \times 10^{-6} \, \text{m} \]
• Order, \( n = 2 \)
• \( \sin 30^\circ = 0.5 \)

Diffraction condition:

\[ n\lambda = d \sin\theta \] \[ \lambda = \frac{d \sin\theta}{n} \] \[ \lambda = \frac{2.5 \times 10^{-6} \times 0.5}{2} \] \[ \lambda = 6.25 \times 10^{-7} \, \text{m} \] \[ \lambda = 6250 \, \text{Å} \]

3. A diffraction grating has 600 lines per millimeter. Light of wavelength 480 nm is incident on it. Calculate the angles for first and second-order maxima.

Given:

• Number of lines, \[ N = 600 \, \text{lines/mm} = 6 \times 10^5 \, \text{lines/m} \]
• Grating spacing, \[ d = \frac{1}{N} \]
• Wavelength, \[ \lambda = 480 \times 10^{-9} \, \text{m} \]

Diffraction equation:

\[ d \sin\theta = n\lambda \] \[ \sin\theta = nN\lambda \]

First-order maximum (n = 1):

\[ \sin\theta_1 = 1 \times 6 \times 10^5 \times 480 \times 10^{-9} \] \[ \theta_1 = \sin^{-1}(0.288) = 16.73^\circ \]

Second-order maximum (n = 2):

\[ \sin\theta_2 = 2 \times 6 \times 10^5 \times 480 \times 10^{-9} \] \[ \theta_2 = \sin^{-1}(0.576) = 35.16^\circ \]