JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 25)
A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10 th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10 th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be ________ mm .
Answer
11
Explanation
In Young's double slit experiment, the distance of $n^{th}$ maxima from central maxima is given by,
$Y=\frac{n\lambda D}{d}$
Here, $n,D$ & $d$ are same for both the cases.
So, $$Y \propto \lambda $$
$$ \Rightarrow {{{Y_1}} \over {{Y_2}}} = {{{\lambda _1}} \over {{\lambda _2}}} \Rightarrow {{10} \over {{Y_2}}} = {{600} \over {660}} = {1 \over {11}}$$
$$ \Rightarrow {Y_2} = 11\,mm$$.
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