Find the velocity of the image of the given o | vedclass

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(B) Given: Focal length $f = +10 \ cm$,object distance $u = -30 \ cm$,velocity of object $\vec{V}_O = +4 \hat{i} \ cm/sec$,velocity of lens $\vec{V}_L

Vedclass · Accepted Answer

$2.5$

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(B) Given: Focal length $f = +10 \ cm$,object distance $u = -30 \ cm$,velocity of object $\vec{V}_O = +4 \hat{i} \ cm/sec$,velocity of lens $\vec{V}_L = +2 \hat{i} \ cm/sec$.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} - \frac{1}{-30} = \frac{1}{10} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$.So,$v = +15 \ cm$.Transverse magnification $m = \frac{v}{u} = \frac{15}{-30} = -0.5$.The formula for velocity of image with respect to the lens is $\vec{V}_{I/L} = m^2 \vec{V}_{O/L}$.$\vec{V}_I - \vec{V}_L = m^2 (\vec{V}_O - \vec{V}_L)$.$\vec{V}_I - 2 \hat{i} = (-0.5)^2 (4 \hat{i} - 2 \hat{i}) = 0.25 (2 \hat{i}) = 0.5 \hat{i}$.$\vec{V}_I = 2 \hat{i} + 0.5 \hat{i} = 2.5 \hat{i} \ cm/sec$.

Find the velocity of the image of the given object in $cm/sec$.

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