In a regular YDSE, when thin film of refractive index mu is placed in front of the upper slit then it is observed that the intensity at the central point becomes half of the original intensity. It is also observed that the initial 3^(rd) maxima is now below the central point and the initial `4^(th)` minima is above the central point. Now, a film of refractive index `mu_(1)` and thickness same as the above film. is put in the front of the lower slit also. It is observed that whole fringe pattern shifts by one fringe width. What is the value of `mu_(1)` ?

Options

A. `(4mu+9)//12`
B. `(4mu+9)//13`
C. `(4mu+9)//11`
D. None

Answer

B

Explanation

`Deltaphi=(2pi)/lambdat(mu-1)....(1)`
`6piltDeltaphilt7pi.......(2)`
`I=I_(0)cos^(2)((Deltaphi)/2)`
`I=I_(0)//2`
`Rightarrow cos((Deltaphi)/2)`
`I=I_(0)//2`
`Rightarrow cos ((Deltaphi)/2)=pm 1/sqrt(2)Rightarrow (Deltaphi)/2=2npipm pi/4`
`Deltaphi=4npipmpi//2.....(3)`
Solving equation (2)&(3)we get
`Delta phi = 6.5pi`
`from (1) 6.5pi-(2pi)/lambda t(mu-1)....(4)`
Introducing record film shifted the fringed pattern by one fringe with means.
`t(mu-1)=lambda....(5)`
Solving 4 & 5
`Rightarrow mu_(1)=(4mu+9)/13`