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Distance between screen and source is decreased by 25%. Then the percentage change in fringe width isA. `20%`B. `31%`C. `75%`D. `25%` |
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Answer» Correct Answer - D `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%` |
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