In Young's double slit experiment, the fringe width is found to be 0.6 mm. without distrubing anything, the whole arrangment is dipped in water of refractive index `4/3` the new fringe width will be

Options

A. 0.15 mm
B. 0.3 mm
C. 0.6 mm
D. 0.45 mm

Answer

D

Explanation

`mu=(lamda_(a))/(lamda_(w))=(4)/(3)" "therefore lamda_(w)=((3)/(4))lamda_(a)`
`because X=(lamdaD)/(d)" "therefore X=lamda`
`therefore X_(a)=(3)/(4)" "X_(air)=(3)/(4)xx0.6mm`
`therefore X_(w)=0.45mm`
Thus fringe width decreases in water.