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

C

Explanation

Path difference for nth order bright fringes,
`d sin theta=(lambda)/(d)=(nlambda)/(2lambda) " " [ :. D=2lambda]`
`:. sin teta=(n)/(2)`
but range of `sin theta` is(-1,1)
`:. Sin theta le1`
`:. n le2`
`:.` Possible values of `n=-2,1,0,1,2`
Hence, a central maximum on the screen with two sides of the first order and two of the second order respectively gives a total of five bright fringes.