CBSE Class 12 Physics 2023 Delhi Set 3 Paper

Question : 6 of 14

Marks: +1, -0

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 the fifth bright fringe at the same point ?

Solution:

Since,

y_n=\;{n λ D}/{d}

For

3^{\;\text "rd "\;}

bright fringe,

y_3=\;{3 × 700 D}/{d}=\;{2100 D}/{d}

For

5^{\;\text "th "\;}

bright fringe,

y_5=5 λ^{'} \;{D}/{d}

Since

y_5=y_3

5 λ' \;{D}/{d}=2100 \;{D}/{d}

∴

λ^{'}=\;{2100}/{5}=420 {\nm}

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