1.

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

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`


Discussion

No Comment Found

Related InterviewSolutions