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JEE Advanced · Physics · 25. Wave Optics

In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of \(0.8 \mathrm{~mm}\). The distance between the slits at time \(t\) is given by \(d=(0.8+0.04 \sin \omega t) \mathrm{mm}\), where \(\omega=0.08 \mathrm{rad} \mathrm{s}^{-1}\). The distance of the screen from the slits is \(1 \mathrm{~m}\) and the wavelength of the light used to illuminate the slits is \(6000 Ã…\). The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point \(O\).

The maximum speed in \(\mu \mathrm{m} / \mathrm{s}\) at which the \(8^{\text {th }}\) bright fringe will move is ________ .

  1. A 24
  2. B 25
  3. C 26
  4. D 27
Verified Solution

Answer & Solution

Correct Answer

(A) 24

Step-by-step Solution

Detailed explanation

\(\begin{aligned}
& \mathrm{y}=\mathrm{n} \cdot \frac{\lambda \mathrm{D}}{\mathrm{d}} \\
& \mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}=-\mathrm{n} \cdot \frac{\lambda \cdot \mathrm{d}}{\mathrm{d}^2} \cdot \frac{d(d)}{d t} \\
& d=0.8+0.04 \sin \omega t \\
& \frac{d(d)}{d t}=0.04 \omega \cos \omega t
\end{aligned}\)
\(\text { for } \mathrm{v} \rightarrow \max \Rightarrow \frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}} \rightarrow \max \text {. }\)
For \(\frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}} \rightarrow \max\).
\(\cos \omega \mathrm{t}=1 \Rightarrow \sin \omega \mathrm{t}=0 \)
\( \Rightarrow\left(\frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}}\right)_{\max }=0.04 \)
\( \Rightarrow \mathrm{d}=0.8 \mathrm{~mm} \)
\( \mathrm{v}_{\max }=\frac{8 \times 6000 \times 10^{-10} \times 1 \times 0.04 \times 0.08}{0.8 \times 0.8 \times 10^{-6} \times 10^{-3}}\) \(=24 \mu \mathrm{m} / \mathrm{s}.\)
(range is from 23.5 to 24.5)
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