A screen is placed `50 cm` from a single slit, which is illuminated with `6000 Å` light. If the distance between the first and third minima in the diffraction pattern is `3.00 mm`, what is the width of the slit ?

Options

A. `1 xx 10^(-4)m`
B. `2xx10^(-4)m`
C. `0.5xx10^(-4)m`
D. `4xx10^(-4)m`

Answer

B

Explanation

The position of nth minima in the diffraction pattern, is
`x_(n)=(n D lambda)/(d)`
`therefore x_(3)-x_(1)=(3-1)(D lambda)/(d)=(2D lambda)/(d)`
So, width of the slit is
`d=(2D lambda)/(x_(3)-x_(1))`
`=(2xx0.50xx6000xx10^(-10))/(3xx10^(-3))`
`=2xx10^(-4) m`