Question

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

• A. $2.20\text{mm}$
• B. $2.64\text{mm}$
• C. $1.10\text{mm}$
• D. $0.5\text{mm}$

Answer 1

The correct option is B $2.64\text{mm}$


Given:-
Wavelength of light, $\lambda =440\text{nm}=440\times {10}^{-9}\text{m}$

$SO=20.0\text{cm}$

$\theta =0.5\times {10}^{-3}\text{rad (very small)}$

$S$ and $S_1$ are source of $YDSE$

$D=SO\cos \theta +100$

$\Rightarrow D=(20\times 1)+100=120\text{cm}(\cos \theta \approx 1)$

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

$\Rightarrow d=2\times {10}^{-2}\text{cm}$

$\therefore$ Fringe width,

$\beta =\frac{\lambda D}{d}=\frac{440\times {10}^{-9}\times 120\times {10}^{-2}}{2\times {10}^{-2}\times {10}^{-2}}$

$=264\times {10}^{-2}\text{mm}=2.64\text{mm}$

Hence, option $\left(\text{B}\right)$ is correct.