A point object is moving at a constant speed | vedclass

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(C) Given: Focal length $f = 10 \,cm$, speed of object $v_o = |du/dt| = 1 \,ms^{-1}$, speed of image $v_i = |dv/dt| = 1 \,ms^{-1}$.From the lens formu

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$20$

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(C) Given: Focal length $f = 10 \,cm$, speed of object $v_o = |du/dt| = 1 \,ms^{-1}$, speed of image $v_i = |dv/dt| = 1 \,ms^{-1}$.From the lens formula: $1/f = 1/v - 1/u$.Differentiating with respect to time $t$: $0 = -1/v^2 (dv/dt) + 1/u^2 (du/dt)$.This gives: $dv/dt = (v^2/u^2) (du/dt)$.Since the speed of the image is equal to the speed of the object, $|dv/dt| = |du/dt|$, we have $v^2/u^2 = 1$, which implies $|v| = |u|$.For a real image formed by a convex lens, the magnification $m = v/u = -1$ (since the image is inverted and real).Using the lens formula $1/f = 1/v - 1/u$, substitute $v = -u$:$1/f = 1/(-u) - 1/u = -2/u$.$u = -2f = -2(10) = -20 \,cm$.The distance from the optical centre is $20 \,cm$.

$A$ point object is moving at a constant speed of $1 \,ms^{-1}$ along the principal axis of a convex lens of focal length $10 \,cm$. The speed of the image is also $1 \,ms^{-1}$, when the object is at what distance (in $cm$) from the optical centre of the lens?

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