Question

In a single slit diffraction experiment first minima for `lambda_1 = 660nm` coincides with first maxima for wavelength `lambda_2`. Calculate the value of `lambda_2`.
A. 240 nm
B. 345 nm
C. 440 nm
D. 330 nm

Answer 1

Correct Answer - C
Given, `lambda=660nm`
Position of minima in diffraction pattern is given by
`a sin theta=n lambda`
For the first minimaof `lambda_(1)`, we get
`a sin theta_(1)=1xx lambda_(1) or sin theta_(1)=(lambda_(1))/(a)`
For the first maxima approximately of wavelength `lambda_(2)`,
`a sin theta_(2)=(3)/(2)lambda_(2), sin theta_(2)=(3lambda_(2))/(2a)`
The two will be considered if, `theta_(1)=theta_(2) or sin theta_(1)=sin theta_(2)`
`(lambda_(1))/(a) =(3lambda_(2))/(2a) or lambda_(2)=(2)/(3)lambda_(1)=(2)/(3)xx660 nm`
`rArr lambda_(2)=440nm`