In YDSE performed with wavelength `lambda = 5890 Å` the angular fringe width is `0.40^(@)`. What is the angular fringe width if the entire set-up is immersed in water?

Explanation

Angular fringe width is given by
`beta_(theta)=(lambda)/(d)`
So, `beta_(theta)^("air")=(lambda_("air"))/(d)`,
`beta_(theta)^("water")=(lambda_("water"))/(d)`
`(beta_(theta)^("air"))/(beta_(theta)^("water"))=(lambda_("air"))/(lambda_("water"))=(n_("air"))/(n_("water"))=(3)/(4)`
Thus `beta_(theta)^("water")=(3)/(4)beta_(theta)^("air")=0.40^(@)xx(3)/(4)=0.30^(@)`