Question

Light of wavelength $\lambda$ is incident on a slit of width $a$. The resulting diffraction pattern is observed on a screen at a distance $D$. The linear width of the principal maximum is equal to the width of the slit, if $D$ equals,

• A. $\frac{a^2}{2\lambda }$
• B. $\frac{a}{\lambda }$
• C. $\frac{2{\lambda }^2}{a}$
• D. $\frac{2\lambda }{a}$

Answer 1

The correct option is A $\frac{a^2}{2\lambda }$

The linear width of central maximum is given by,
$y=2\beta =\frac{2\lambda D}{a}$
If it is equal to width of slit $a$, then,
$\frac{2\lambda D}{a}=a$

$\therefore D=\frac{a^2}{2\lambda }$

Hence, $\left(A\right)$ is the correct answer.