Question

A single slit of width $b$ is illuminated by a coherent monochromatic light of wavelength $\lambda .$ If the second and fourth minima in the diffraction pattern at a distance $1cm$ from the slit are at $3cm$ and $6cm$ respectively from the central maximum, what is the width of the central maximum? (i.e. distance between first minimum on either side of the central maximum)

• A. $4.5cm$
• B. $1.5cm$
• C. $6.0cm$
• D. $3.0cm$

Answer 1

Answer: (D)

For secondary minima $dsin\theta =n\lambda$
$sin\theta =\frac{n\lambda }{d}$
For second minima
$n=2$
$sin\theta =\frac{2\lambda }{d}=tan{\theta }_1=\frac{x_1}{D}$
For fourth minima $n=4$
$sin{\theta }_2=\frac{4\lambda }{d}=\frac{x_2}{D}$
Distance between both minima is $x_2-x_1=\frac{4\lambda }{d}-\frac{2\lambda }{d}=\frac{2\lambda }{d}=6-3=3cm$
Width of central maxima, $=\frac{2\lambda }{d}=3\text{c}\text{m}$