Question

In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

• A. 0.5 mm
• B. 0.02 mm
• C. 0.2 mm
• D. 0.1mm

Answer 1

The Correct Option is C

For double slit experiment 
\(d=1 mm = 1\times {10}^{-3} m , D=1m , \lambda = 500 \times {10}^{-9} m\) 
Fringe width \(\beta = \frac{ S \lambda}{ d}\) 
Width of central maxima in a single slit 
As per question, width of central maxima of single 
slit pattern = width of 10 maxima of double slit 
pattern 
\(\frac{ 2 \lambda D}{a} = 10 \bigg( \frac{\lambda D}{ d} \bigg)\)
\(a= \frac{ 2d}{10} = \frac{2 \times {10}^{-3}}{ 10}\)
\(= 0.2 \times {10}^{-3} m\)
\(= 0.2 mm\)
Therefore, the correct option is (C) : 0.2 mm

Answer 2

d = 10^-3m
D = 1m
λ = 500×10^-9m
The width of central maxima in a single slit diffraction:  \(\frac{2\lambda D}{a}\)
Fringe width in double slit pattern: β= \(\frac{\lambda D}{d}\)
Given,
​10β= \(\frac{2\lambda D}{a}\)
⇒10 \(\frac{\lambda D}{d}\)= \(\frac{2\lambda D}{a}\)
⇒ a= \(\frac{d}{5}\) mm
⇒  0.2mm
Therefore, the correct option is (C) : 0.2 mm