Distance between screen and source is decreased by 25%. Then the percentage change in fringe width is

Options

A. `20%`
B. `31%`
C. `75%`
D. `25%`

Answer

D

Explanation

`D_(1)-D_(12)=25%, lambda_(1)-beta_(2)=?`
`D_(1)-D_(2)=(25)/(100)=(1)/(4)`
`:. beta prop D`
`(beta_(2))/(beta_(1))=(D_(2))/(D_(1))=(3)/(4)(D)/(D)`
`beta_(2)=3(3beta_(1))/(4)`
`beta_(1)-beta_(2)=beta_(1)(3 beta_(1))/(4)=(beta_(1))/(4)`
% change `=((beta_(1)-beta_(1))/(beta_(1)))100=(beta_(1))/( 4xx beta_(1))xx 100=25%`