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प्रश्न
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?
विकल्प
420 nm
750 nm
630 nm
500 nm
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उत्तर
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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