JEE MainPhysicsEasyWave Optics

The distance between two slits in a Young's double slit
experiment is 3 mm. The distance of the screenfrom the slits is 1 m.
Waves of wavelength 1 mm are incident on the plane of the slits
normally.The distance of the first maxima on the screen from the central
maxima will be: (in cm)

Show answer and solution

At first maxima d sinθ = λ

3sin\theta = 1 \Rightarrow sin\theta = \frac{1}{3}

\tan\theta = \frac{y}{D} \Rightarrow y = D \times \tan\theta = 1 \times \frac{1}{\sqrt{8}}m = \frac{100}{\sqrt{8}}cm = 35.4cm

Source: Grass-G / P-20 Physcis Paper 12 Dec.docx

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