Wave Optics

JEE Main level, numerical answer, medium

137 of 191

Question 137 of 191

Sodium light of wavelengths 650nm650nm and 655nm655nm is used to study diffraction at a single slit of aperture 0.5mm0.5mm. The distance between the slit and the screen is 2.0m2.0m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is _____ ×10−5m\times {10}^{-5}m

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Answer: 3

The condition for maxima in diffraction is,
bsin⁡θ=(2n+1)λ2b\sin \theta =\left(2n+1\right)\dfrac{\lambda }{2}.
For the first maxima n=1n=1. Therefore,
bsin⁡θ=3λ2⇒sin⁡θ=3λ2b=yDy=3λD2bb\sin \theta =\dfrac{3\lambda }{2} \Rightarrow \sin \theta =\dfrac{3\lambda }{2b}=\dfrac{y}{D} y=\dfrac{3\lambda D}{2b}
Now the difference between the positions of the maxima will be,
y1−y2=3D2b(λ1−λ2)⇒y1−y2=3×22×5×10−4×(655−650)×10−9=3×10−5m{y}_{1}-{y}_{2}=\dfrac{3D}{2b}\left({\lambda }_{1}-{\lambda }_{2}\right) \Rightarrow {y}_{1}-{y}_{2}=\dfrac{3\times 2}{2\times 5\times {10}^{-4}}\times \left(655-650\right)\times {10}^{-9}=3\times {10}^{-5}m

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