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JEE Advanced · Physics · 25. Wave Optics

Paragraph I: 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 \(8^{\text {th }}\) bright fringe above the point \(\mathrm{O}\) oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer \((\mu \mathrm{m})\), is ___________ .

  1. A 604.5
  2. B 601.5
  3. C 605.7
  4. D 615.8
Verified Solution

Answer & Solution

Correct Answer

(B) 601.5

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & y=n \cdot\left(\frac{\lambda D}{d}\right) \\ & \text { for } 8^{\text {th }} \text { fringe } \\ & y=8 \frac{\lambda D}{d} \\ & y_{\max }=8 \frac{\lambda D}{d_{\min }} \\ & y_{\min }=8 \frac{\lambda D}{d_{\max }} \\ & y_{\max }-y_{\min }=8 \lambda D\left[\frac{1}{d_{\min }}-\frac{1}{d_{\max }}\right] \\ & \lambda=6000 Ã…\end{aligned}\)
\(\mathrm{D}=1 \mathrm{~m} \)
\( \mathrm{~d}_{\max }=0.34 \mathrm{~mm} \)
\( \mathrm{~d}_{\text {min }}=0.76 \mathrm{~mm} \)
\( y_{\max }-y_{\min }=8 \times 6000 \times 10^{-10} \times 1\) \(\left[\frac{1}{0.76 \times 10^{-3}}-\frac{1}{0.84 \times 10^{-3}}\right] \)
\( =8 \times 6 \times 10^{-4} \times\left[\frac{0.08}{0.76 \times 0.84}\right]=601.5 \mu \mathrm{m}\)
(answer range is from 598 to 602 )
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