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

Answer: (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 $