Wave Optics

JEE Main level, numerical answer, medium

75 of 191

Question 75 of 191

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 5×10−2m5\times {10}^{-2}m towards the slits, the change in fringe width is 3×10−3cm3\times {10}^{-3}cm. If the distance between the slits is 1mm1mm, then the wavelength of the light will be____nmnm.

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Answer: 600

We know that the fringe width in YDSE is β=λDd\beta =\dfrac{\lambda D}{d}
⇒Δβ=λdΔD\Rightarrow \Delta \beta =\dfrac{\lambda }{d}\Delta D
⇒λ=(Δβ)dΔD=3×10−3×10−2×10−35×10−2=6×10−7m=600nm\Rightarrow \lambda =\dfrac{\left(\Delta \beta \right)d}{\Delta D}=\dfrac{3\times {10}^{-3}\times {10}^{-2}\times {10}^{-3}}{5\times {10}^{-2}}=6\times {10}^{-7}m=600nm

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