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.

Options

A. `650mn, 6.32nm`
B. `350nm, 4.33nm`
C. `452nm, 3.53nm`
D. `650nm, 433nm`

Answer

D

Explanation

At O : path difference is `Delta x = Delta x_(2) = ((mu_(g))/(mu_(m)) -1)t`
For maximum intensity at O
`D x =n1 (Here n = 1,2,3,…..)`
`1= (Delta x)/(1) , (Delta x)/(2) , (Delta x)/(3)` ….. And so on `Delta x = ((1.5)/(4//3) -1)`
`(10.4xx10^(-6)m) = ((1.5)/(4//3) - 1) (10.4xx10^(3)mm)`
Maximum intensity will be corresponding to
`1 = 13000 nm, (1300)/(2) nm, (1300)/(3)m, (1300)/(4)nm` .....
`= 1300nm, 650 nm, 433.33nm, 325 nm nm` .....
The wavelengths in the range 400 to 700 nm are 650 nm and `433.33` nm.