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Q. The maximum number of possible interference maxima when slit separation is equal to $4$ times the wavelength of light used in a double slit experiment is

KCETKCET 2008Wave Optics

Solution:

$ d \sin \theta= n \lambda $
$ \sin \theta=\frac{ n }{4} $
maximum values of $\sin \theta$ can not exceed \pm 1 So possible value os $n$ is $0, \pm 1, \pm 2, \pm 3, \pm 4$ answer is 9