A particle is moving in a circle of radius $1 c m$ and with constant speed of $8 c m / s$ . Centre of the circle lies on principal axis of a converging lens of focal length $50 c m$ and at a distance of $75 c m$ from the lens. Plane of the circle is perpendicular to principal axis. If speed of image is $2 x c m / s$ then $x$ is :

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Solution

From the laws of magnification, we determine the size of the radius of the circle.
m = $\frac{f}{f + u}$

m = $\frac{50}{50 - 75}$ = -2

So, the radius of the image circle will be, $2 c m$
Now, the angular velocity should remain constant, so

$\frac{8}{1} = \frac{2 x}{2}$

This gives, x = 8.

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From the laws of magnification, we determine the size of the radius of the circle.m = ff+um = 5050−75 = -2So, the radius of the image circle will be, 2cmNow, the angular velocity should remain constant, so81=2x2This gives, x = 8.

From the laws of magnification, we determine the size of the radius of the circle.
m = $\frac{f}{f + u}$

m = $\frac{50}{50 - 75}$ = -2

So, the radius of the image circle will be, $2 c m$
Now, the angular velocity should remain constant, so

$\frac{8}{1} = \frac{2 x}{2}$

This gives, x = 8.