Question

A beam of light of wavelength $500\text{nm}$ from a distant source falls on a single slit $2\text{mm}$ wide and the resulting diffraction pattern is observed on a screen $2.2\text{m}$ away. The distance between the first dark fringes on either side of the central bright fringe is

• A. $1.1\text{cm}$
• B. $1.2\text{cm}$
• C. $2.4\text{cm}$
• D. $1.1\text{mm}$

Answer 1

The correct option is D $1.1\text{mm}$

The distance between the first dark fringes on either side of central maximum is equal to the width of central maximum

$W=\frac{2\lambda D}{a}$

Where
$a\to \text{Width of slit}=2\text{mm}$
$\lambda \to \text{Wavelength of light}=500\text{nm}$
$D\to \text{distance of screen from slit}=2.2\text{m}$

$\Rightarrow W=\frac{2\times 500\times {10}^{-9}\times 2.2}{2\times {10}^{-3}}\text{m}$

$\Rightarrow W=1.1\text{mm}$

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

Why this question? Asked in $\text{IIT -JEE 1994}$.