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, $=\frac{2 \lambda }{d}=3 \, \text{c}\text{m}$