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QJEEM23Phd700: JEE Main 2023 Physics - Class 12 JEEM23Phd700 Solution: Two light waves of wavelengths 800 and are used in Young's…

Two light waves of wavelengths 800 and 600 nm600 \mathrm{~nm} are used in Young's double slit experiment to obtain interference fringes on a screen placed 7 m7 \mathrm{~m} away from plane of slits. If the two slits are separated by 0.35 mm0.35 \mathrm{~mm} , then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______ mm\mathrm{mm} .

Step-by-Step Verified Solution

ω1=λ1Dd\omega_{1}=\frac{\lambda_{1} D}{d} and ω2=λ2Dd\omega_{2}=\frac{\lambda_{2} D}{d}

ω1=16 mm\omega_{1}=16 \mathrm{~mm} and ω2=12 mm\omega_{2}=12 \mathrm{~mm}

So LCM⁡(ω1,ω2)=48 mm\operatorname{LCM}\left(\omega_{1}, \omega_{2}\right)=48 \mathrm{~mm}

So at 48 mm48 \mathrm{~mm} distance both bright fringes will be found.

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