MHT CET · Physics · Wave Optics
In Young's double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits.
If the screen is moved by \(5 \times 10^{-2} \mathrm{~m}\) towards the slits, the change in fringe width is \(3 \times 10^{-5} \mathrm{~m}\). If the separation between the slits is \(10^{-3} \mathrm{~m}\), the wavelength of light used is
- A \(6000 Å\)
- B \(5000 Å\)
- C \(45000 Å\)
- D \(3000 Å\)
Answer & Solution
Correct Answer
(A) \(6000 Å\)
Step-by-step Solution
Detailed explanation
\(\omega=\frac{\lambda D}{d}\)
\(\Rightarrow \Delta \omega=\frac{\lambda(\Delta D)}{d}\)
\(\therefore \lambda=\frac{(d)(\Delta \omega)}{\Delta D}=\frac{\left(\left(10^{-3}\right)\left(3 \times 10^{-5}\right)\right)}{5 \times 10^{-2}}\)\(=0.6 \times 10^{-6} \mathrm{~m}=6000 Å\)
\(\Rightarrow \Delta \omega=\frac{\lambda(\Delta D)}{d}\)
\(\therefore \lambda=\frac{(d)(\Delta \omega)}{\Delta D}=\frac{\left(\left(10^{-3}\right)\left(3 \times 10^{-5}\right)\right)}{5 \times 10^{-2}}\)\(=0.6 \times 10^{-6} \mathrm{~m}=6000 Å\)
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