A YDSE is performed in a medium of refractive index `4 // 3`, A light of 600 nm wavelength is falling on the slits having 0.45 nm separation . The lower slit `S_(2)` is covered b a thin glass plate of thickness 10.4 mm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in figure. (All the wavelengths in this problem are for the given medium of refractive index `4 // 3`, ignore absorption.)

Find the light intensity at point O relative t maximum fringe intensity.

Options

A. `(3)/(2)I_(max)`
B. `(5)/(4)I_(max)`
C. `(3)/(4)I_(max)`
D. `(7)/(5)I_(max)`

Answer

C

Explanation

At `O, Delta x_(1) = 0 and Delta x_(2) = ((mu_(g))/(mu_(m)) -1)t`
Net path difference, `Delta x = Delta x_(2)`
Corresponding phase difference, `Delta phi` or simply `phi = (2pi)/(lambda) . Delta x` Substituting the value, we have
`phi = (2pi)/(6xx10^(-7)) ((1.5)/(4//3)-1) = ((13)/(3))pi`
Now, `I(phi)= I_(max) cos^(2)((phi)/(2)) , I = I_(max)cos^(2) ((13pi)/(6))`
`I = (3)/(4) I_(max)`