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In Young's double slit experiment, the 8...

In Young's double slit experiment, the 8th maximum with wavelength `lambda_1` is at a distance `d_1` from the central maximum and the 6th maximum with a wavelength `lambda_2` is at a distance `d_2`. Then `(d_1//d_2)` is equal to

A

`(4)/(3)((lambda_(2))/(lambda_(1)))`

B

`(4)/(3)((lambda_(1))/(lambda_(2)))`

C

`(3)/(4) ((lambda_(2))/(lambda_(1)))`

D

`(3)/(4) ((lambda_(1))/(lambda_(2)))`

Text Solution

Verified by Experts

The correct Answer is:
B
Position of nth maxima from central maxima is
given by ` x_(n) = (n lambda D)/(d)`
for 8th maxima `x_(8) = (8 lambda_(1) D)/(d_(1))`
and for 6th maxima `x_(6) = (6 lambda_(2) D)/(d_(2))`
Now, ` x_(6) = x_(8) , rArr (d_(1))/(d_(2)) = (n_(1)lambda_(1))/(n_(2)lambda_(2)) = (4)/(4) ((lambda_(1))/(lambda_(2)))`
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