Wave Optics

JEE Main level, numerical answer, medium

46 of 191

Question 46 of 191

A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10 th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be _________ mm .

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Answer: 11

Position of the nth n^{\text {th }} bright fringe w.r.t. central maxima
in a YDSE is yn=nλDdy_n=n \frac{\lambda D}{d}.
⇒y10′y10=λ′λ\Rightarrow \quad \frac{y_{10}^{\prime}}{y_{10}}=\frac{\lambda^{\prime}}{\lambda}
or y10′=λ′λy10\quad y_{10}^{\prime}=\frac{\lambda^{\prime}}{\lambda} y_{10}
=(660 nm600 nm)10 mm=11 mm\begin{aligned} & =\left(\frac{660 \mathrm{~nm}}{600 \mathrm{~nm}}\right) 10 \mathrm{~mm} \\ & =11 \mathrm{~mm}\end{aligned}

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