In a Young's double slit interference experiment the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength `lamda`. Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one fourth of the maximum.

Answer

A, B, C, D

Explanation

i. When intensity is half the maximum,
`rarr I/I_(max)=1/2`
`rarr (4a^2cos^2(pi/2))/(4a^2)=1/2`
`rarr cos^2(phi/2)=1/2`
`rarr cos(phi/2)=1/sqrt2`
`rarr phi/2=pi/4`
`rarr path difference
`x=1/4`
`rarr y=(xD)/d=(lamdaD)/(4d)`
ii When intensity is the maximum
`I/I_(max)=1/4`
`implies 4a^2cos^2(phi/2)=1/4`
`rarr cos^2(phi/2)=1/4`
`rarr cos(phi/2)=1/2`
`rarr phi/2=pi/3`
`rarr path difference x=lamda/3`
`rarr y=(xD)/d=(lamdaD)/(3d)