In a YDSE using monochromatic visible light, the distance between the plate of slits and the screen is 1.7 m. At point P on the screen which is directly in front of the upper slit, maximum path is observed. Now, the screen is moved 50 cm closer to the plane of slits. Point P now lies between third and fouth minima above the central maxima and the intensity at P in one-fourth of the maxima intensity on the screen.
Find the value of n.

Options

A. 4
B. 6
C. 2
D. 8

Answer

c

Explanation

For point P in front of upper slit, `y = d // 2`.
Initially, path diffence,
`((d // 2) d)/(D) = n lambda`
Intensity at any point on the screen is
`I = I_(max) ((1 - cos phi)/(2))`
(`phi` being the phase difference)
When the screen is placed, `I = (I_(max))/(4)`
Hence, `cos phi = (1)/(2) implies phi = 2 n pi +- (pi)/(3)`
Also, as P lie between third and forth minima,
`5 pi lt phi lt 7pi`
From Eqs. (ii) and (iii), we get
`phi = (17 pi)/(3)` or `(19 pi)/(3)`
`phi = ((d // 2) d)/(D') (2 pi)/(lambda) = ((n lambda D))/(D') (2 pi)/(D') = (2n pi D)/(D')`
`= (17 n pi)/(6)` (putting the values of D and D')
`:. n = 2` or `(38 // 17)`
But n is integer, therefore `n = 2` is the only valid answer.
b. Putting the value of n and d in Eq. (i), we get `lambda = 5.9 xx 10^(-7) m`.