JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 21)
Two light waves of wavelengths 800 and $600 \mathrm{~nm}$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7 \mathrm{~m}$ away from plane of slits. If the two slits are separated by $0.35 \mathrm{~mm}$, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______ $\mathrm{mm}$.
Answer
48
Explanation
$\omega_{1}=\frac{\lambda_{1} D}{d} $ and $ \omega_{2}=\frac{\lambda_{2} D}{d}$
$\omega_{1}=16 \mathrm{~mm} $ and $ \omega_{2}=12 \mathrm{~mm}$
So $\operatorname{LCM}\left(\omega_{1}, \omega_{2}\right)=48 \mathrm{~mm}$
So at $48 \mathrm{~mm}$ distance both bright fringes will be found.
$\omega_{1}=16 \mathrm{~mm} $ and $ \omega_{2}=12 \mathrm{~mm}$
So $\operatorname{LCM}\left(\omega_{1}, \omega_{2}\right)=48 \mathrm{~mm}$
So at $48 \mathrm{~mm}$ distance both bright fringes will be found.
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