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  • The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double - slit experiment is:Option: 1</strong

The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double - slit experiment is:

Option: 1

infinite


Option: 2

five


Option: 3

three


Option: 4

zero


Answers (1)

best_answer

As \mathrm{\mathrm{d} \sin \theta=\mathrm{n} \lambda} and \mathrm{\mathrm{d}=2 \lambda, 2 \sin \theta=\mathrm{n}} and as such maximum value of \mathrm{ \mathrm{n}=2}. Thus, interference maxima are formed when\mathrm{ \mathrm{n}=0, \mathrm{n}= \pm 1, \mathrm{n}= \pm 2}, i.e., maximum number of possible interference maxima is 5 .

Posted by

Suraj Bhandari

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