Question

In a single slit diffraction experiment, the aperture of the slit is 3 mm and the separation between the slit and the screen is 1.5 m. A monochromatic light of wavelength 600 nm is normally incident on the slit. Calculate the distance of (I) first order minimum, and (II) second order maximum, from the centre of the screen.

Hint:

In single-slit diffraction, minima occur at \( a \sin \theta = m\lambda \), and secondary maxima lie roughly midway between them. Use small angle approximation \( \sin \theta \approx \frac{y}{D} \) for small diffraction angles.

Answer 1

Given: - Slit width \( a = 3\,\text{mm} = 3 \times 10^{-3}\,\text{m} \) - Distance to screen \( D = 1.5\,\text{m} \) - Wavelength \( \lambda = 600\,\text{nm} = 600 \times 10^{-9}\,\text{m} \)\underline{(I) First Order Minimum:}
In single slit diffraction, minima are located at:\[a \sin \theta = m\lambda \quad \text{for } m = \pm 1, \pm 2, \ldots\]For small angles, \( \sin \theta \approx \tan \theta = \frac{y}{D} \). Substituting this into the equation for minima gives:\[a \cdot \frac{y_1}{D} = \lambda \Rightarrow y_1 = \frac{\lambda D}{a}= \frac{600 \times 10^{-9} \times 1.5}{3 \times 10^{-3}} = 3 \times 10^{-4}\,\text{m} = 0.3\,\text{mm}\]Distance of first order minimum = \( \boxed{0.3\,\text{mm}} \)\underline{(II) Second Order Maximum (Approximate):}
Secondary maxima in single slit diffraction are not sharp and are located approximately halfway between successive minima.Therefore, the second order maximum is positioned roughly between the 1st and 2nd minima.The position of the 2nd minimum is calculated as:\[\text{Position of 2nd minimum: } y_2 = \frac{2\lambda D}{a}= \frac{2 \times 600 \times 10^{-9} \times 1.5}{3 \times 10^{-3}} = 0.6\,\text{mm}\]The approximate position of the second order maximum is:\[\text{Approximate position of 2nd maximum: } y \approx \frac{y_1 + y_2}{2} = \frac{0.3 + 0.6}{2} = 0.45\,\text{mm}\]Distance of second order maximum \( \approx \boxed{0.45\,\text{mm}} \)