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 lamda " " `… (i)
Given , ` d = ` slit - width = ` 2lamda `
` therefore 2lamda sin theta = n lamda `
` rArr 2 sin theta = n `
The maximum value of` sin theta ` is 1, hence,
` n = 2 xx 1 = 2 `
Thus, Eq. (i) must be satisfied by 5 integer values i.e. ` - 2, - 1, 0, 1, 2 `.
Hence, the maximum number of possible interferrence maxima is 5.