question_answer

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is                                 [AIEEE 2004]

A) infinite

B) five

C) three

D) zero

Correct Answer: B

Solution :

[b] For possible interference maxima on the screen, the condition is

...(i)

Given,

The maximum value ofis 1, hence

Thus, Eq. (i) must be satisfied by 5 integer values i.e., -2, -1, 0, 1, 2. Hence, the maximum number of possible interference maxima is 5.