A plane wave of wavelength `6250Å` is incident normally on a slit of width `2xx10^(-2)` cm The width of the principal maximum of diffraction pattern on a screen at a distance of 50 cm will be

Options

A. `312.5xx10^(-3) cm`
B. `312.5xx10^(-4)cm`
C. `312 cm`
D. `312.5xx10^(-5) cm`

Answer

A

Explanation

According to the question
`lambda=6250 Å=6250xx10^(-10)xx10^(2)cm=6250xx10^(-8)cm`
Width of the slit `a=2xx10^(-2)cm`
The angular width of central maxima `beta=2theta =2lambda/a`
Thus, linear width is
`L=betaxxD=(2lambda)/axxD=(2xx6250xx10^(-8))/(2xx10^(-2))xx50`
`=6250xx10^(-6)xx50=3125xx10^(-4)`
`=312.5xx10^(-3)cm`