In Young's double slit experiment, the distance between two slits is made three times then the fringe width will becomes

Options

A. `I_(0)/4`
B. `3/4 I_(0)`
C. `I_(0)/2`
D. `I_(0)`

Answer

C

Explanation

In the above figure `S_(1)` and `S_(2)` are the two different slits Given, distance between slits `S_(1)` and `S_(2) d=5lambda`
distance between screen and slits `D=10d=50 lambda`
Here, `lambda` is the wavelength of light used in the experiment According to the question the intensity at maximum in this Young's double slit experiment is `I_(0)`
`rArr I_(max)=I_(0)`
`:.` Path difference `=(dY_(n))/D=(d xx(d//2))/(10d)=d/20=lambda/4[because d=5lambda]`
A path difference of `lambda` corresponding to phase difference `2pi` So, for path difference `lambda//4` phase difference
`phi=(2pi)/lambdaxxlambda/4=pi//2=90^(@)`
As we have `I=I_(0)"cos"^(2)phi/2rArrI=I_(0)"cos"^(2)90^(@)/2`
`rArrI=I_(0)xx(1/sqrt2)^(2)rArrI=I_(0)/2`