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In a single slit diffraction experiment ...

In a single slit diffraction experiment the first minimum for red light of wavelength `6600Å` coincides with the first maximum for other light of wavelength `lamda`. The value of `lamda` is

A

4400 Å

B

6600 Å

C

2000 Å

D

3500 Å

Text Solution

Verified by Experts

The correct Answer is:
A
In a single slit diffraction experiment, position of minima is given by `d sin theta=n lambda`
So, for first minima of red `sin theta=1 xx ((lambda_(R))/(d))` and as first maxima is midway between first and second minima, for wavelength `lambda` its position will be `d sin theta'=(lambda+2lambda)/(2) rArr sin theta' =(3lambda')/(2d)`
According to given condition `sin theta=sin theta' rArr lambda'=2/3 lambdaR`
So, `lambda' =2/3 xx 6600=440nm =4400 Å`
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