Question

In a double slit experiment, the two slits are 1 mm apart and the screen is place 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?

Answer 1

Double slit experiment

1. A distinctive pattern of bright and dark fringes are visible when monochromatic light passes through two narrow openings and superimpose on the screen.
2. The superposition of overlapping light waves originating from the two slits causes this interference pattern.

Step 1: Given data

Monochromatic light of wavelength $\lambda =500nm=500\times {10}^{-9}m[1nm={10}^{-9}m]$

The two slits are 1 mm apart and the screen is placed 1 m away i.e., $d={10}^{-3}m,D=1m$

Step 2 : Formula used

Fringe width in double-slit pattern: $\beta =\frac{\lambda D}{d}$

Step 3: Calculating the width of the fringe

$$\begin{gathered} \beta =\frac{\lambda D}{d} \\ \Rightarrow \beta =\frac{500\times {10}^{-9}\times 1}{{10}^{-3}}=500\times {10}^{-6}m=0.5mm \end{gathered}$$

Hence, the width of the fringe is $0.5mm$