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

For possible interference maxima on the screen the condition is
`d sin theta = n lambda`
Given : d = slit - width `= 2lambda`
`therefore 2 lambda sin theta = n lambda rArr 2 sin theta = n` eq …….(i)
The maximum value of `sin theta` is 1 hence.
`n = 2 xx 1 = 2`
Thus, Eq. (i) must be satisfied by 5 integer values ie, -2-1, -1, 2. Hence the maximum number of possible interference maxima is 5.