1.

if the intensity of the principal maximum in the single slit Fraunhoffer diffraction pattern, then the intensity when the slit width is double will be .....

Answer»

`I_(0)`
`(I_(0))/(2)`
`2I_(0)`
`4I_(0)`

Solution :The intensity on the screen by angle `theta`
`I_(theta)=I_(0)=(sin^(2) alpha)/(alpha) "" [ "where" alpha=(pi d sin theta)/(lambda)]`
`:.` For central maximum `alpha=(pi(2d)sin theta^(0))/(lambda)`
`alpha=0`
`:. I_(theta)=I_(0)=(lim_(alpha to 0)(sin alpha)/(alpha))^(2)`
`=I_(0)"" [ :. lim_(alpha to 0)(sin alpha)/(alpha)=1]`
Short method : Maximum intensity depend on source of light. Here source is unchanged hence maximum intensity ALSO cannot CHANGED.


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