The bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer (m), is [2024]

Question

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 $d = (0.8 + 0.04 \sin ω t) mm,$ where $ω = 0.08 {rad s}^{- 1}$ . 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. [2024]

Q. The $8^{th}$ bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer ( $μ$ m), is ______.

Smart Achievers · Accepted Answer

(601.50)

Fringe width for the $d = (0.8 + 0.04 \sin ω t) mm,$ fringe $y = n (\frac{λ D}{d})$

For the $8^{th}$ fringe, $y = 8 \frac{λ D}{d}$

$6000 Å$ and $y_{\min} = 8 \frac{λ D}{d_{\max}}$

$y_{\max} - y_{\min} = 8 λ D [\frac{1}{d_{\min}} - \frac{1}{d_{\max}}]$

Given, $D = 1 m, λ = 6000 \times 10^{- 10} m$

$d_{\max} = 0.84 mm$

$d_{\min} = 0.76 mm$

$∴$ $y_{\max} - y_{\min} = 8 \times 6000 \times 10^{- 10} \times 1 [\frac{1}{0.76 \times 10^{- 3}} - \frac{1}{0.84 \times 10^{- 3}}]$

$= 8 \times 6 \times 10^{- 4} [\frac{0.08}{0.76 \times 0.84}]$ $= 6.015 \times 10^{- 4} m = 601.5 μ m$

Solution

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