Question

In a Young's double slit experiment, two slits are located 1.5 m apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is $ x \times 10^{-3} \, \text{cm} $, where x-value is:

Hint:

In diffraction experiments, the distance between slits and the wavelength of light can be used to calculate the dimensions of the slits.

Answer 1

Correct Answer: 15

The formula for the angular position of maxima in double slit diffraction is: \[ \frac{d}{a} = 2\theta \] where \( d = 1.5 \, \text{m} \) (distance between the slits) and \( a \) is the width of the slit. For the single slit diffraction pattern, the angular position of the first minima is: \[ \frac{2D}{a} = \frac{400}{400 \, \text{nm}} = 15 \times 10^{-3} \, \text{cm} \]
Thus, the width of each slit is \( 15 \times 10^{-3} \, \text{cm} \).