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QJEEM22Ph5d42: JEE Main 2022 Physics - Class 12 JEEM22Ph5d42 Solution: In Young's double slit experiment the two slits are 0.6 mm…

In Young's double slit experiment the two slits are 0.6 mm distance apart. Interference pattern is observed on a screen at a distance 80 cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ____________ nm.

Step-by-Step Verified Solution

y=d2y = {d \over 2} ,

∴\therefore Δx=ydD\Delta x = y{d \over D}

⇒d22D=λ2\Rightarrow {{{d^2}} \over {2D}} = {\lambda \over 2}

⇒λ=(0.6×10−3)20.8\Rightarrow \lambda = {{{{(0.6 \times {{10}^{ - 3}})}^2}} \over {0.8}}

= 450 nm

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