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 Young's Double Slit Experiment (YDSE) configuration, the provided parameters are:- Slit separation, \( d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \),- Distance from slits to screen, \( L = 2 \, \text{m} \),- Light wavelength, \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \),- The number of maxima observed within the central maximum of the single-slit diffraction pattern is 20.For a double-slit setup, the separation between adjacent maxima is calculated using:\[y_{\text{max}} = \frac{\lambda L}{d}\]where \( y_{\text{max}} \) denotes the distance between consecutive maxima.In single-slit diffraction, the angular width of the central maximum is defined as:\[\theta_{\text{central}} = \frac{\lambda}{a}\]with \( a \) representing the width of each individual slit.The distance between adjacent minima in a single-slit diffraction pattern is given by:\[y_{\text{min}} = \frac{\lambda L}{a}\]It is stated that 20 double-slit maxima fall within the central maximum of the single-slit diffraction. This implies:\[20 \times y_{\text{max}} = y_{\text{min}}\]Substituting the derived expressions for \( y_{\text{max}} \) and \( y_{\text{min}} \):\[20 \times \frac{\lambda L}{d} = \frac{\lambda L}{a}\]Simplifying the equation yields:\[20 \times \frac{1}{d} = \frac{1}{a}\]\[a = 20d\]Using the given value for \( d = 1.5 \times 10^{-3} \, \text{m} \):\[a = 20 \times 1.5 \times 10^{-3} = 3 \times 10^{-2} \, \text{m} = 2 \, \text{mm}\]Consequently, the width of each slit is determined to be \( 2 \, \mu\text{m} \).Therefore, the correct answer is (3) 2 μm.