Monochromatic light of wavelength `5000 Å` is used in YDSE, with slit width, `d = 1 mm`, distance between screen and slits, `D = 1 M`. If intensites at the two slits are `I_1 = 4I_0 and I_2 = I_0`, find:
a. finge width `beta:`
b. distance of 5th minima from the central maxima on the screen,
c. intensity at `y = (1)/(3) mm,`
d. distance of the 1000th maxima, and
e. distance of the 5000th maxima.

Explanation

a. `beta = (lambda D)/(d) = (5000 xx 10^(-10) xx 1)/1 xx 10^(-3) = 0.5 mm`
b. ` y = (2n - 1) (lambda D)/(D), n = 5 implies y = 2.25 mm`
c. At `y = (1)/(3) mm, y lt lt D implies Delta x = (dy)/(D)`
`(Delta phi) = (2 pi)(dy)/(lambda) Delta x = 2 pi (dy)/(lambda D) = (4 pi)/(3)`
Now , resultant intensity,
`I = I_(1) +I_(2 )+ 2 cos (Delta phi)= 4I_(0) + I_(0) + 2 sqrt( 4 I_0^(2)) cos (Delta phi)`
`5 I_(0 )+ 4 I_(0) cos ((4 pi)/(3)) = 3 I_(0)`
d. `(d)/(lambda) = (10^(3))/(0.5 xx 10^(6)) = 2000`
Hence, now `Delta x = d sin theta` must be used.
` :. d sin theta = nlambda = 1000 lambda implies sin theta = 1000 (lambda)/(d) = (1)/(2)`
` implies theta = 30^(@)`
` y = D tan theta = (1)/(sqrt 3) m`
e. Highest order maxima, `n_max = [(d)/(lambda)] = 2000`
Hence, `n = 5000` is not possible.