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 1 m from the slit are at $3$ cm and $6$ cm 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. $6.0cm$
• C. $1.5cm$
• D. $3.0cm$

Answer 1

The correct option is C $1.5cm$

Difference between second and first minimum $=6-3=3$

Which is also given as $x_y-x_2$

Where $x_n=\frac{D}{d}\left[\right(2n-1\left)\frac{\lambda }{2}\right]$

$\therefore x_y=\frac{D}{d}\left(\frac{7\lambda }{2}\right),x_2=\frac{3\lambda D}{2d}$

$\Rightarrow x_y-x_2=3cm=\frac{D}{d}(\frac{7\lambda }{2}-\frac{3\lambda }{2})$

$\Rightarrow 3cn=2\lambda \frac{D}{d}\Rightarrow \frac{D\lambda }{d}=\frac{3}{2}=1.5cm$

Now, width of central maximum $=2\times x_1$

$=2\times \frac{D}{d}\left[\right(2\times 1-1\left)\frac{\lambda }{2}\right]$

$=\frac{D\lambda }{d}$

$=1.5cm$