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Question
In an interference experiment, a 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 the fifth bright fringe at the same point?
Options
420 nm
750 nm
630 nm
500 nm
Solution
420 nm
Explanation:
Position of nth bright fringe,
`y_n = (nlambdaD)/d`
Given: λ = 700 nm
∴ `y_3 = (3 xx 700 xx D)/d`
⇒ `y_3 = (2100D)/d`
The 5th bright fringe due to light of wavelength λ' is formed at y3,
∴ y5 = y3
or `(5 xx lambda^'D)/d = (2100 nmD)/d`
⇒ `lambda^' = 2100/5` = 420 nm
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