Question

In a YDSE setup, the slits are separated by 1.5 mm and the distance between the slits and the screen is 2 m. On using light of wavelength 400 nm, it is observed that 20 maxima of double slit experiment lie inside the central maxima of single slit diffraction. The width of each slit is μm.

• A. 0.5 μm
• B. 1 μm
• C. 2 μm
• D. 3 μm

Hint:

The number of maxima in a double-slit diffraction pattern within the central maximum of a single-slit diffraction pattern can help determine the slit width using the relationship between the maxima and minima in the diffraction patterns.

Answer 1

The Correct Option is C

In this YDSE setup, we are given the following parameters: - The distance between the slits is \( d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \), - The distance between the slits and the screen is \( L = 2 \, \text{m} \), - The wavelength of the light is \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \), - The number of maxima observed inside the central maximum of single slit diffraction is 20. In a double-slit diffraction, the distance between adjacent maxima is given by the formula: \[ y_{\text{max}} = \frac{\lambda L}{d} \] where: - \( y_{\text{max}} \) is the distance between adjacent maxima. For the single-slit diffraction, the angular width of the central maximum is given by: \[ \theta_{\text{central}} = \frac{\lambda}{a} \] where \( a \) is the width of each slit. The distance between adjacent minima in single-slit diffraction is: \[ y_{\text{min}} = \frac{\lambda L}{a} \] Now, we are told that 20 maxima of the double-slit diffraction lie within the central maxima of the single-slit diffraction. Therefore: \[ 20 \times y_{\text{max}} = y_{\text{min}} \] Substituting the expressions for \( y_{\text{max}} \) and \( y_{\text{min}} \): \[ 20 \times \frac{\lambda L}{d} = \frac{\lambda L}{a} \] Simplifying this equation: \[ 20 \times \frac{1}{d} = \frac{1}{a} \] \[ a = 20d \] Substitute \( d = 1.5 \times 10^{-3} \, \text{m} \): \[ a = 20 \times 1.5 \times 10^{-3} = 3 \times 10^{-2} \, \text{m} = 2 \, \text{mm} \] Thus, the width of each slit is \( 2 \, \mu\text{m} \). Therefore, the correct answer is (3) 2 μm.