JEE Advanced 2024 Paper 2 · Q14 · 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 mm. The distance between the slits at time $t$ is given by $d = (0.8 + 0.04\sin\omega t)$ mm, where $\omega = 0.08$ rad s$^{-1}$. The distance of the screen from the slits is 1 m and the wavelength of the light used to illuminate the slits is 6000 Å. 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 O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer (μm), is _____.
Reveal answer + step-by-step solution
Correct answer:601.50
Solution
Position of $n^{\text{th}}$ bright fringe: $y_n = n\lambda D/d$.
For $n = 8$: $y_8 = 8\lambda D/d$.
Extreme positions occur at extreme values of $d$: $d_{\min} = 0.8 - 0.04 = 0.76$ mm $= 0.76\times 10^{-3}$ m, $d_{\max} = 0.8 + 0.04 = 0.84$ mm $= 0.84\times 10^{-3}$ m.
With $\lambda = 6000$ Å $= 6\times 10^{-7}$ m, $D = 1$ m:
$y_{8,\max} = \dfrac{8\cdot 6\times 10^{-7}\cdot 1}{0.76\times 10^{-3}} = \dfrac{4.8\times 10^{-6}}{0.76\times 10^{-3}} = 6.3158\times 10^{-3}$ m.
$y_{8,\min} = \dfrac{8\cdot 6\times 10^{-7}\cdot 1}{0.84\times 10^{-3}} = \dfrac{4.8\times 10^{-6}}{0.84\times 10^{-3}} = 5.7143\times 10^{-3}$ m.
Separation $\Delta y = (6.3158 - 5.7143)\times 10^{-3}$ m $= 0.6015\times 10^{-3}$ m $= 601.5$ μm.
Units: $\lambda$, $D$, $d$ all in m; $y$ in m, then converted to μm. Answer: $601.50$.
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