Discussion Forum



1)

In Fraunhofer diffraction pattern, slit width is 0.2 mm and the screen is at 2m away from the lens,If wavelength of light used is 5000Å , then the distance between  the first minimum on either  side of the central maximum is 

($\theta$  is small and measured in radian)


A) $10^{-3}$ m

B) $10^{-2}$ m

C) $2 \times 10^{-3}$ m

D) $2 \times 10^{-1}$ m

Answer:

Option B

Explanation:

Given , a =0.2 x 10-3 m, D=2 m

 $\lambda$  = 5 x 10-7 m

 As,   $y=\frac{\lambda D}{a}=\frac{5\times 10^{-7}\times2}{0.2\times 10^{-3}}$

 $\therefore$    $y=\frac{5\times 10^{-7}}{ 10^{-4}}$

$\therefore$    $y=5\times 10^{-3}$ m

Distance between 1 st minima on either side

 $=5\times 10^{-3}+5\times 10^{-3}=10\times 10^{-3}=10^{-2}m$