Wave Optics

JEE Main level, single correct, medium

64 of 191

Question 64 of 191

The width of fringe is 2mm2mm on the screen in a double slit experiment for the light of wavelength of 400nm400nm. The width of the fringe for the light of wavelength 600nm600nm will be:

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Answer: (D) 3mm3mm

For a light of wavelength λ\lambda, the fringe width in double slit experiment is given by
β=λDd...(1)\beta =\dfrac{\lambda D}{d}...(1)
For a light of wavelength λ′\lambda ', the fringe width is given by
β′=λ′Dd...(2)\beta '=\dfrac{\lambda 'D}{d}...\left(2\right)
Divide equation (2) by equation (1) and simplify to obtain the fringe width for the later case.
β′β=λ′DdλDd=λ′λ⇒β′=λ′λβ...(3)\begin{matrix}\dfrac{\beta '}{\beta } & = & \dfrac{\dfrac{\lambda 'D}{d}}{\dfrac{\lambda D}{d}} \\ & = & \dfrac{\lambda '}{\lambda } \\ & \Rightarrow & \beta '=\dfrac{\lambda '}{\lambda }\beta ...\left(3\right)\end{matrix}
Substitute the values of the known parameters into equation (3) to calculate the required fringe width.
β′=600nm400nm×2mm=3mm\begin{matrix}\beta ' & = & \dfrac{600nm}{400nm}\times 2mm \\ & = & 3mm\end{matrix}

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