Wave Optics

JEE Main level, numerical answer, medium

71 of 191

Question 71 of 191

In a Young's double slit experiment, a laser light of 560nm560nm produces an interference pattern with consecutive bright fringes' separation of 7.2mm7.2mm. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1mm8.1mm. The wavelength of second light is _____ nm.nm.

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Answer: 630

Given here: λ1=560×10−9m{\lambda }_{1}=560\times {10}^{-9}m, β1=7.2×10−3m{\beta }_{1}=7.2\times {10}^{-3}m and β2=8.1×10−3m{\beta }_{2}=8.1\times {10}^{-3}m.
Let dd be the width and DD be the distance between slit and screen then fringe width, β=λDd\beta =\dfrac{\lambda D}{d}.
Thus, from above, β∝λ\beta \propto \lambda
So, the wavelength of second light is λ2=β2β1λ1=98λ1{\lambda }_{2}=\dfrac{{\beta }_{2}}{{\beta }_{1}}{\lambda }_{1}=\dfrac{9}{8}{\lambda }_{1}
∴λ2=98×560=630nm.∴{\lambda }_{2}=\dfrac{9}{8}\times 560=630nm.

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