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In an interference experiment, third bright fringe is obtained at a point on the screen with a light of $700 \mathrm{~nm}$. What should be the wavelength of the light source in order to obtain 5 th bright fringe at the same point?
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Verified Answer
The correct answer is:
$420 \mathrm{~nm}$
$n_{1} \lambda_{1}=n_{2} \lambda_{2}$
$$
\Rightarrow \quad 3 \times 700=5 \times \lambda_{2}
$$
$$
\Rightarrow \quad \lambda_{2}=420 \mathrm{~nm}
$$
$$
\Rightarrow \quad 3 \times 700=5 \times \lambda_{2}
$$
$$
\Rightarrow \quad \lambda_{2}=420 \mathrm{~nm}
$$
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