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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 - Physics

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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

MCQ

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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2022-2023 (March) Delhi Set 3

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