Question

A monochromatic light source S of wavelength $440$ nm is placed slightly above a plane mirror M as shown. 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 $20.0$cm, and the distance of screen from O is $100.0$ cm (figure is not to scale). If the angle $\theta =0.50\times {10}^{-2}$ radians, 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

The correct option is B $2.64$ mm

Given:

Wavelength, $\lambda =44nm=440\times {10}^{-9}m$

$SO=20.0cm$

$\theta =0.5\times {10}^{-2}radians$(very small)

To find:

Width of fringes

S and S${}_1$ are source of YDSE

$D=SO\cos \theta +100=20\times 1+100=120cm$

$d=2\times SO\sin \theta =2\times 20\times 0.5\times {10}^{-3}=2\times {10}^{-2}cm$

$\beta =\frac{\lambda D}{d}=\frac{440\times {10}^{-6}\times 120\times {10}^2}{2\times {10}^{-2}\times {10}^2}=264\times {10}^{-2}=2.64mm$

is the required fringe width.