A single slit of width a is illuminated by violet light of wavelength `400nm` and the width of the diffraction pattern is measured as y. When half of the slit width is covered and illuminated by yellow light of wavelength `600nm`, the width of the diffraction pattern is

Options

A. 0
B. `(y)/(3)`
C. 3y
D. 4y

Answer

C

Explanation

Given, `lambda_(1)=4000 nm, lambda_(2)=600 nm, d_(1)=d and d_(2)=(d)/(2)`
In single slit diffraction experiment, width of central maxima
`y=(2lambdaD)/(d) rArr (y_(1))/(y_(2))=(lambda_(1))/(lambda_(2))xx(d_(2))/(d_(1))`
`therefore (y_(1))/(y_(2))=(400)/(600)xx((d)/(2))/(d)=(1)/(3)`
So, `y_(2)=3y_(1)=3y`