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In an interference experiment, third bright fringe is obtained at a point on the screen with a light of 700 nm . What should be the wavelength of the light source in order to obtain 5th bright fringe at the same pointA. 630 mmB. 500 mmC. 420 mmD. 750 mm |
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Answer» Correct Answer - C `lamda_(1)=700nm,n_(1)=3n_(2)=5,lamda=?` For same path diff. `n_(1)lamda_(1)=n_(2)lamda_(2)` `lamda_(2)=(n_(1)lamda_(1))/(n_(2))=(3xx700)/(5)=420nm` |
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