JEE MAIN - Physics (2022 - 29th June Evening Shift - No. 26)
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
600
Explanation
Fringe width $$\beta = {{\lambda D} \over d}$$
$$ \Rightarrow \left| {d\beta } \right| = {\lambda \over d}\left| {d(D)} \right|$$
$$ \Rightarrow 3 \times {10^{ - 3}}\,cm = {\lambda \over {1\,mm}}\left( {5 \times {{10}^{ - 2}}\,m} \right)$$
$$ \Rightarrow \lambda = {{3 \times {{10}^{ - 8}}} \over {5 \times {{10}^{ - 2}}}}\,m$$
$$ \Rightarrow \lambda = 600\,nm$$
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