An object AB of height 1cm lying on the axis of convex mirror of focal length20cm, at a distance of 30cm from its pole as shown in figure. Now the object starts moving with a constant speed `1 cm//sec` towards the pole. Calculate

Q. Rate of change of height of image.

Options

A. `0.16cms^(-1)`
B. `0.004cms^(-1)`
C. `0.016cms^(-1)`
D. `0.0008cms^(-1)`

Answer

d.

Explanation

(i) Given : `u=-30cm, f=20cm`
`(1)/(v)+(1)/(u)=(1)/(f)rArrv=12cm`
Also, `(dv)/(dt)=-((v^(2))/(u^(2)))(du)/(dt)=-0.16cms^(-1)`
(ii) `H=-(v)/(u)h`, H is the height of image and h is height of object.
`(dH)/(dt)=(u(dv)/(dt)-v(du)/(dt))/(u^(2))h=0.008cms^(-1)`