JEE MAIN - Physics (2004 - No. 2)
The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is :
three
five
infinite
zero
Explicació
For constructive interference $$d\,\sin \theta = n\lambda $$
Given $$d = 2\lambda \Rightarrow \sin \theta = {n \over 2}$$
$$n = 0,1, - 1,2, - 2$$ hence five maxima are possible.
Given $$d = 2\lambda \Rightarrow \sin \theta = {n \over 2}$$
$$n = 0,1, - 1,2, - 2$$ hence five maxima are possible.