The max. No. Of possible interference maxima for slit separation equal to twice the wavelength of YDSE is :

The max. No. Of possible interference maxima for slit separation equal

1.	The max. No. Of possible interference maxima for slit separation equal to twice the wavelength of YDSE is :
Answer» `INFTY` 5 3 zero Solution :d sin `THETA = n lambda` `d = 2 lambda` `thereforesin theta = n/2` value of sin `theta` VARIES between - 1 and + 1.So equation (i) is true if `n = -1, -2, 0, +1, +2.` `therfore` n has five integer values.

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The max. No. Of possible interference maxima for slit separation equal to twice the wavelength of YDSE is :

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