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In YDSE monochromatic light of wavelengt...

In YDSE monochromatic light of wavelength 600 nm incident of slits as shown in figure.

If `S_(1)S_(2) = 3mm, OP=11 mm` then

A

If `alpha = (0.36)/(pi)` degree then destructive interface at point P

B

If `alpha = (0.36)/(pi)` degree then constructive interfaces at point O

C

If `alpha = 0` then constructive interface at O

D

Fringe width depends an `alpha`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C

`Delta x = d sin alpha + dsin theta`
`theta alpha` : small angle
`sin theta ~=tan theta = (y)/(b)`
`Delta x = d alpha + (d y)/(D)`
(A) `Delta x = 3mm xx (0.36)/(pi)xx(pi)/(180)+(3mmxx11mm)/(1)=39000 `
`3900 n, = (2n -1) (lambda)/(2)=(2n-1)xx(600nm)/(2)`
`n = 7`
destructive inference happened
(B) `Delta x = 3mm xx(0.36)/(pi)xx(pi)/(180)+0=600 nm`
`600 nm = n lambda`
`n =1`
constructive interfrence
(C ) `Delta x = 0`
So constructive interference
(D) Fringe width does not depend on `alpha`.
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