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.
Answer (integer)
450
Solution
<p>$y = {d \over 2}$,</p>
<p>$\therefore$ $\Delta x = y{d \over D}$</p>
<p>$\Rightarrow {{{d^2}} \over {2D}} = {\lambda \over 2}$</p>
<p>$\Rightarrow \lambda = {{{{(0.6 \times {{10}^{ - 3}})}^2}} \over {0.8}}$</p>
<p>= 450 nm</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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