Question

In a single slit diffraction pattern, a light of wavelength \( 6000 \, \text{Å} \) is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from the slits. The width of the slit is ______ \( \times 10^{-4} \) m.

Answer 1

Correct Answer: 2

The width of a slit in a single slit diffraction pattern is determined using the formula for minima positions: \(a \sin \theta = n\lambda\). Here, \(a\) represents slit width, \(\theta\) is the diffraction angle, \(n\) is the order of the minima, and \(\lambda\) is the light's wavelength. For small angles, \(\sin \theta \approx \tan \theta = \frac{y}{D}\), where \(y\) is the distance of the minima from the central maximum on the screen, and \(D\) is the slit-to-screen distance.

The distance between the first (\(n=1\)) and third (\(n=3\)) minima is 3 mm. The positions of these minima are given by \(y_1 = \frac{\lambda D}{a}\) and \(y_3 = \frac{3\lambda D}{a}\).

The difference in these positions is: \(y_3 - y_1 = \frac{3\lambda D}{a} - \frac{\lambda D}{a} = \frac{2\lambda D}{a}\).

This difference is equated to the given distance: \(\frac{2\lambda D}{a} = 3 \, \text{mm} = 0.003 \, \text{m}\).

Substituting \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m}\) and \(D = 50 \, \text{cm} = 0.5 \, \text{m}\):

\(a = \frac{2 \cdot 6000 \times 10^{-10} \cdot 0.5}{0.003}\).

\(a = \frac{6000 \times 10^{-10}}{0.003} = 2 \times 10^{-4} \, \text{m}\).

Therefore, the slit width is \(2 \times 10^{-4} \, \text{m}\), which falls within the specified range [2,2].