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. The patternn vanishes and the width is zero
B. `y/3`
C. `3y`
D. None of these

Answer

C

Explanation

given: `lamda=400nm and lamda'=600 mm and a'=a/2`
the width of the central maximum of the diffraction
pattern is `(2lamdaD)/(a)=y and y'=(2lamda'D)/(a//2)`
`therefore (y')/(y)=(2lamda'D)/(a//2)xx(a)/(2lamda D)=(lamda')/(lamda)xx2=2xx(600)/(400)=3`
`therefore y'=3y`