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# A point is moving on the principal axis of a concave mirror of focal length 24cm towards the mirror When it is at distance of 60cm from the mirror its velocity is 9cm/sec.what is the velocity of the image at that instant 11 years ago

1/f = 1/u + 1/v

f=24 cm , u=60 cm on substituting above gives

so, v = 40 cm

also :  1/u2.(du/dt) =  -1/v2.(dv/dt)

1600*9 = -3600*dv/dt

velocity of image = 4 cm/sec towards mirror

--

regards

Ramesh

one year ago

## Weknowf1​=u1​+v1​;wheref=focallength=24cmu=initialdistance=60cmv=finaldistance=?241​−601​=v1​v=40cmAlsou21​dtdu​=v21​dtdv​dtdu​=initialvelocity=9cm/sanddtdv​=finalvelocity=vv=36009×1600​=4cm/s

one year ago
\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\\    -(1)
So using this formula we can have the v i.e. distance of image as :
\frac{1}{v} = \frac{-1}{24}+\frac{-1}{60} \\
On solving this we get v = -40 cm
On differentiating eq 1 we get :
\frac{-1}{v^{2}} .\frac{dv}{dt} - \frac{1}{u^{2} }. \frac{du}{dt} = o\\
O