Q. The velocity of image w.r.t ground in the below figure is
(a) 45 m/s and approaches the mirror
(b) 45 m/s and moves away from the mirror
(c) 60 m/s and approaches the mirror
(d) 60 m/s and moves away from the mirror
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Ans: (a)
Sol: According to mirror formula ,
$ \displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $
Differentiating
$ \displaystyle -\frac{dv}{v^2} – \frac{du}{u^2} = 0 $
$ \displaystyle \frac{dv}{du} = -\frac{v^2}{u^2} $
Dividing by dt ,
$ \displaystyle \frac{dv/dt}{du/dt} = -\frac{v^2}{u^2} $
$ \displaystyle \frac{V_I}{V_O} = -(\frac{60}{20})^2 = -9 $
VI = -9Vo = -9 × 5 = -45 m/s