First slide

Diffraction

Question

In Fraunhoffer diffraction pattern width of central bright fringe is $β$ . If wavelength of light is doubled and slit width is halved, distance of second dark fringe from the centre of central bright fringe is

Difficult

A

4 $β$
B

2 $β$
C

2.5 $β$
D

3 $β$

Solution

If ‘a’ be the width of the slit and $λ$ be the wavelength of light used. Then $β = \frac{2 λ}{a}$ .
In the second case width of central maxima is
$β^{1} = \frac{2 λ^{1}}{a^{1}} = \frac{2 (2 λ)}{(\frac{a}{2})} = 4 (\frac{2 λ}{a}) = 4 β$
$∴$ Width of first secondary maxima $= \frac{β^{1}}{2} = 2 β$
$∴$ Distance of second dark fringe
= Half of the width of central bright fringe + width of first secondary bright fringe
$= \frac{β^{1}}{2} + \frac{β^{1}}{2} = β^{1} = 4 β$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App