Question

A monochromatic light source $S$ of wavelength $440nm$ is placed slightly above a plane mirror $M$ as shown below. Image of $S$ in $M$ can be used as a virtual source to produce interference fringes on the screen. The distance of source $S$ from $O$ is $20.0cm$ and the distance of screen from $O$ is $100.0cm$ (figure is not to scale). If the angle $0=0.50$ x $10^{-3}$ radians , then the width of the interference fringes observed on the screen is

• A. 2.20 mm
• B. 2.64 mm
• C. 1.10 mm
• D. 0.55 mm

Answer 1

Answer: (B)

Given arrangement is

Angle $\theta =0.5\times 10^{-3}$radians.
SO, distance between source $S$ and its image $S_1$ is
$d=2r\theta =2\times SO\times \theta$
$=2\times 20\times 0.5\times 10^{-3}$cm
$=20\times 10^{-3}cm=2\times 10^{-4}m$
Distance of sources $S$ and $S_1$ from screen
$D=a+b=20+100=120cm$
$=120\times 10^{-2}m$
Wavelength of light used is
$\lambda =440nm=440\times 10^{-9}m$
Both $S$ and $S_1$ acts like slits (sources of coherent light) and produces fringes over screen.
Width of a fringe, $\beta =\frac{\lambda D}{d}$
$\therefore \beta =\frac{440\times 10^{-9}\times 120\times 10^{-2}}{C}$
$=2.64mm$