In a double slit experiment the angular width of a fringe is found to be `0.2^(@)` on a screen placed I m away. The wavelength of light used in 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water ? Take refractive index of water to be `4//3`.

Options

A. `0.15^(@)`
B. `1^(@)`
C. `2^(@)`
D. `0.3^(@)`

Answer

A

Explanation

Since a fringe of width `beta` is formed on the screen at distance D from the slits, so the angular fringe width
`theta=(beta)/(D)=(D lambda//d)/(D)=(lambda)/(d) " " [:' beta=(D lambda)/(d)]`
` implies d=(lambda)/(theta)`
If the wavelength in water be `lambda'` and the angular fringe width be `theta'`, then
`d=(lambda')/(theta')"or"(lambda)/(theta)=(lambda')/(theta') " or "theta'=(lambda')/(lambda).theta=(lambda//mu)/(lambda).theta[ :' lambda'=(lambda)/(mu)]`
` :. theta'=(0.2^(@))/(4//3)=0.15^(@)`