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Grade: 12

                        

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

Answers : (3)

Ramesh V
70 Points
							

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

11 years ago
ankit singh
askIITians Faculty
596 Points
							

Weknowf1=u1+v1;wheref=focallength=24cmu=initialdistance=60cmv=finaldistance=?241601=v1v=40cmAlsou21dtdu=v21dtdvdtdu=initialvelocity=9cm/sanddtdv=finalvelocity=vv=36009×1600=4cm/s

ASK
2 months ago
ankit singh
askIITians Faculty
596 Points
							
\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
2 months ago
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