In a YDSE experiment, one of the slits is covered with a glass plate of thickness (2.4×10−4) meter, refractive index of which varies with time as μ=t2+12t+8. The separation between the slits is 2×10−3 meter and the distance between the plane of slits and the screen is 2 meters. Find the velocity (in m/s) of the central maxima at t = 4 s. (Take refractive index of air = 1).(Round off to nearest integer)

(μ−1)t0=d×yDy=(μ−1)t0Dd⇒velocity of central maxima, v=dydt=(2t+12)×2.4×10−4×22×10−3at t = 4 secv=4.8 m/s

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In a YDSE experiment, one of the slits is covered with a glass plate of thickness $(2.4 \times 10^{- 4})$ meter, refractive index of which varies with time as $μ = t^{2} + 12 t + 8.$ The separation between the slits is $2 \times 10^{- 3}$ meter and the distance between the plane of slits and the screen is 2 meters. Find the velocity (in m/s) of the central maxima at t = 4 s. (Take refractive index of air = 1).
(Round off to nearest integer)

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Detailed Solution

$(μ - 1) t_{0} = \frac{d \times y}{D}$

$y = (μ - 1) t_{0} \frac{D}{d} \Rightarrow velocity of central maxima,$

$v = \frac{d y}{d t} = (2 t + 12) \times \frac{2.4 \times 10^{- 4} \times 2}{2 \times 10^{- 3}}$

$at t = 4 sec$

$v = 4.8 m / s$

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