JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2018
- Medium
- Numerical
The angular width of the central maximum in a single slit diffraction pattern is \(60^o \) . The width of the slit is \(1\ \mu m\). The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at distance \(50\ cm\) from the slits. If the observed fringe width is \(1\ cm\), what is slit separation distance?.........\(\mu m\)
(i.e. distance between the centres of each slit. )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(25\)
Step-by-step Solution
Detailed explanation
Angular width of central maxima \(=\frac{2 \lambda}{\mathrm{d}}\)
or, \(\lambda=\frac{\mathrm{d}}{2} ;\) Fringe width, \(\beta=\frac{\lambda \times \mathrm{D}}{\mathrm{d}^{\prime}}\)
\(10^{-2}=\frac{d}{2} \times \frac{50 \times 10^{-2}}{d^{\prime}}=\frac{10^{-6} \times 50 \times 10^{-2}}{2 \times d^{\prime}}\)
Therefore, slit separation distance, \(\mathrm{d}^{\prime}=25 \,\mu \mathrm{m}\)
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