In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5 $\times$ 10$-$2 m towards the slits, the change in fringe width is 3 $\times$ 10$-$3 cm. If the distance between the slits is 1 mm, then the wavelength of the light will be ____________ nm.
Answer (integer)
600
Solution
<p>Fringe width $\beta = {{\lambda D} \over d}$</p>
<p>$\Rightarrow \left| {d\beta } \right| = {\lambda \over d}\left| {d(D)} \right|$</p>
<p>$$ \Rightarrow 3 \times {10^{ - 3}}\,cm = {\lambda \over {1\,mm}}\left( {5 \times {{10}^{ - 2}}\,m} \right)$$</p>
<p>$$ \Rightarrow \lambda = {{3 \times {{10}^{ - 8}}} \over {5 \times {{10}^{ - 2}}}}\,m$$</p>
<p>$\Rightarrow \lambda = 600\,nm$</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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