Question

A single slit of width $b$ is illuminated by a coherent monochromatic light of wavelength $\lambda$. If the second and the fourth minima in the diffraction pattern at a distance $1\text{m}$ from the slit are at $3\text{cm}$ and $6\text{cm}$ respectively from the central maximum, what is the width of the central maximum?

• A. $1.5\text{cm}$
• B. $3.0\text{cm}$
• C. $4.5\text{cm}$
• D. $6.0\text{cm}$

Answer 1

The correct option is B $3.0\text{cm}$

Distance from center at which n'th minima forms is
$y_n=\frac{n\lambda D}{d}$
Given second minima as 3.0 cm,
So,
$y_2=\frac{2\lambda D}{d}$
$y_2=\frac{2\lambda D}{d}=3.0cm$
We Know,$\sin \theta =\frac{\lambda }{d}$
$\frac{y}{D}=\frac{\lambda }{d}$
$y=\frac{\lambda D}{d}$
From diagram the width of central maxima is 2y, i.e., $y_2$.