Wave Optics

JEE Main level, numerical answer, medium

79 of 191

Question 79 of 191

In Young's double slit experiment the two slits are 0.6mm0.6mm distance apart. Interference pattern is observed on a screen at a distance 80cm80cm 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 _____ nmnm.

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Answer: 450

According to the question, the first dark fringe is observed directly opposite to one of the slits. Therefore, half of the fringe width is equal to half of the distance between the slits.
β2=d2⇒β=d=6mm\dfrac{\beta }{2}=\dfrac{d}{2}\Rightarrow \beta =d=6mm
⇒λDd=0.6×10−3⇒λ=0.6×10−3dD⇒λ=0.6×10−3×0.6×10−380×10−2\Rightarrow \dfrac{\lambda D}{d}=0.6\times {10}^{-3}\Rightarrow \lambda =\dfrac{0.6\times {10}^{-3}d}{D}\Rightarrow \lambda =\dfrac{0.6\times {10}^{-3}\times 0.6\times {10}^{-3}}{80\times {10}^{-2}}
⇒λ=450×10−9m=450nm\Rightarrow \lambda =450\times {10}^{-9}m=450nm

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