The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is

Options

A. infinite
B. five
C. three
D. zero

Answer

b

Explanation

The conditon of interferece maxima is
`dsintheta =nlambdarArrsintheta=(nlambda)/(d)`
`Given, d=2lambdarArrsintheta=(nlambda)/(2lambda)=n//2`
The magnitude of sin theta lies between 0 and 1
when `n=0,sintheta=0rArrtheta=0`
When `n=1, sintheta=0//2rArrtheta=30^(@)`
When `n=2, sintheta=1rArr theta=90^(@)`
Thus there is central maximum `(theta=0^(@)` on ther side of it maxima lie at `theta=30^(@) and theta =90^(@)`, so maximum number of possible interference maxima is 5