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 .....

Options

A. `I_(0)`
B. `(I_(0))/(2)`
C. `2I_(0)`
D. `4I_(0)`

Answer

A

Explanation

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.