The maximum speed in at which the bright fringe will move 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 maximum speed in $μ m/s$ at which the $8^{th}$ bright fringe will move is _______.

Smart Achievers · Accepted Answer

(24)

Speed, $d = (0.8 + 0.04 \sin ω t) mm,$ $6000 Å$

$d = 0.8 + 0.04 \sin ω t$ $(given)$

$\frac{d (d)}{d t} = 0.04 ω \cos ω t$

For $V_{\max}$ , $\frac{d (d)}{d t} ⟶ \max$

For $\frac{d (d)}{d t} ⟶ \max,$

$\cos ω t = 1$ $\Rightarrow \sin ω t = 0$

$\Rightarrow {(\frac{d (d)}{d t})}_{\max} = 0.04 ω$

$\Rightarrow d = 0.8 mm$

$∴$ $V_{\max} = \frac{8 \times 6000 \times 10^{- 10} \times 1 \times 0.04 \times 0.08}{0.8 \times 0.8 \times 10^{- 6} \times 10^{- 3}}$ $= 24 μ m/s .$

Solution

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Solution

Ans. (24) Speed, v=dydt=-n·λDd2·d(d)dt [∵ y=nλDd] d=0.8+0.04sinωt (given) d(d)dt=0.04ω cos ωt For Vmax, d(d)dt⟶max For d(d)dt⟶max, cos ωt=1⇒sin ωt=0 ⇒(d(d)dt)max=0.04ω ⇒d=0.8 mm ∴ Vmax=8×6000×10-10×1×0.04×0.080.8×0.8×10-6×10-3=24 μm/s.

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