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.)

Now, if 600 nm, find the wavelength of the ligth that forms maximum exactly at point O.

Explanation

(a) `mu_(g)=3//2, mu_(m)=4//3, ._(m)mu_(g)=mu_(g)/mu_(m)=(3//2)/(4//3)=9/8`,
`D=1.5 m, lambda=600 nm`
`(._(m)mu_(g)-1)t=(yd)/D`
`(9/8-1)(10.4xx10^(-6))=(yxx0.45xx10^(-3))/1.5`
`y=4.33xx10^(-3) m=4.33 mm`
(b) `Deltax=(._(m)mu_(g)-1)t=(9/8-1)xx1.4xx10^(-6)=1.3xx10^(-6) m`
`phi=(2pi)/lambda Deltax=(2pi)/(600xx10^(-9))xx1.3xx10^(-6)=4.33 pi=4pi+pi/3`
`I_(R)=I_(0) cos^(2) (phi//2)=I_(0) cos^(2)(2pi+pi//6)`
`=I_(0) cos^(2) (pi//6)=3/4I_(0)`
(c ) `Deltax=(._(m)mu_(g)-1)t=nlambdaimplies1.3xx10^(-6)=nlambda`
`lambda=1300/n nm`
`n=1, lambda=1300 nm`
`n=2, lambda=650 nm`
`n=3, lambda=433 nm`
`n=4, lambda=325 nm`
In the given range `(400-700 nm)`, wavelength are `650 nm` and `433 nm`.