Question

A light having wavelength 6400 \( \text{\AA} \) is incident normally on a slit of width 2 mm. Then the linear width of the central maximum on the screen kept 2 m from the slit is:

• A. 2.4 cm
• B. 1.28 mm
• C. 1.28 cm
• D. 2.4 mm

Hint:

In diffraction, the linear width of the central maximum can be calculated using the formula \( W = 2L \times \theta \), where \( \theta \) is the angular width of the central maximum.

Answer 1

The Correct Option is B

This is a question related to single-slit diffraction. In single-slit diffraction, the angular width of the central maximum is given by the formula: \[ \theta = \frac{\lambda}{a} \] where: - \( \lambda \) is the wavelength of the light, - \( a \) is the width of the slit. Given: - The wavelength of the light is \( \lambda = 6400 \, \text{\AA} = 6400 \times 10^{-10} \, \text{m} \), - The width of the slit is \( a = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \). Substituting the values into the equation for \( \theta \): \[ \theta = \frac{6400 \times 10^{-10}}{2 \times 10^{-3}} = 3.2 \times 10^{-4} \, \text{radians} \] Next, the linear width of the central maximum on the screen is given by: \[ W = 2L \times \theta \] where: - \( L = 2 \, \text{m} \) is the distance from the slit to the screen. Substituting the values: \[ W = 2 \times 2 \times 10^{-3} \times 3.2 \times 10^{-4} = 1.28 \times 10^{-3} \, \text{m} = 1.28 \, \text{mm} \] Thus, the linear width of the central maximum on the screen is \( 1.28 \, \text{mm} \). Therefore, the correct answer is (B) 1.28 mm.