First slide

Diffraction

Question

In a single slit diffraction experiment first minimum for red light (660 nm) coincides with first maximum of some other wavelength $λ^{'} .$ The value of $λ^{'}$ is

Moderate

A

$4400 Å$
B

$6600 Å$
C

$2000 Å$
D

$3500 Å$

Solution

In a single slit diffraction experiment, position of minima is given by d $\sin θ = n λ$
so for first minima of red $sinθ = 1 \times (\frac{λ_{R}}{d})$ and as first maxima is midway between first and second minima, for wavelength $λ^{'},$ its position will be ${dsinθ}^{'} = \frac{λ^{'} + 2 λ^{'}}{2} \Rightarrow {sinθ}^{'} = \frac{3 λ^{'}}{2 d}$
According to given condition $\sin θ = \sin θ^{'}$ $\Rightarrow λ^{'} = \frac{2}{3} λ_{R}$
$so λ^{'} = \frac{2}{3} \times 6600 = 440 nm = 4400 Å$

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