1.

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

Answer»

8
4
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9

Solution :The width of the fringe is `beta=(lambdaD)/(d)`
Given that `d=4lambda,beta=(lambdaD)/(4lambda)=(D)/(4)`.
NORMALLY D is the order of metres (0.5 to 1M?) `beta` is very LARGE .
The number of orders that can be accomodated is less than one for normal screen . No option given.


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