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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics3 Marks
Question
In Young's double$-$slit experiment the slits are $2\ mm$ apart and interference is observed on a screen placed at a distance of $100\ cm$ from the slits. It is found that the ninth bright fringe is at a distance of $2.208\ mm$ from the second dark fringe from the centre of the fringe pattern. Find the wavelength of light used.
✓

Answer

$d =2\ mm=2 \times 10^{-3} m$,
$D=100 \ cm=1 m$,
$y_9-y_2^{\prime}=2.208\ mm=2.208 \times 10^{-3} m ($on the same side of the centre of the fringe pattern$)$
$y_n=\frac{n \lambda D}{d} ... ($bright fringe$)$
$y_m^{\prime}=\frac{(2 m-1) \lambda D}{2 d} ... ($dark fringe$)$
$\therefore y_9-y_2^{\prime}=\frac{9 \lambda D}{d}-\frac{3 \lambda D}{2 d}=\frac{15 \lambda D}{2 d}$
$\therefore$ The wavelength of light,
$\lambda=\frac{2 d\left(y_9-y_2^{\prime}\right)}{15 D}$
$=\frac{2 \times 2 \times 10^{-3} \times 2.208 \times 10^{-3}}{15 \times 1}$
$=5.888 \times 10^{-7} m$
$=5888 \mathring A $

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