Light of wavelength `589.3nm` is incident normally on the slit of width `0.1mm`. What will be the angular width of the central diffraction maximum at a distance of `1m` from the slit?

Options

A. `0.68^(@)`
B. `0.34^(@)`
C. `2.05^(@)`
D. None of these

Answer

B

Explanation

`beta=(lambdaD)/d=(589.3xx10^(-9))/(0.1xx10^(-3))D`
Hence, angular width of the central diffraction
`theta=beta/D=(589.3xx10^(-9))/(0.1xx10^(-3))"rad"`
`=589.3xx10^(-5)xx180/pi=0.34^(@)`