The speed at which the image of a luminous po | vedclass

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Question

(D) The mirror formula is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Since $f = \frac{r}{2}$,we have $\frac{1}{v} + \frac{1}{u} = \frac{2}{r}$

Vedclass · Accepted Answer

$v_1 = -v_0 \left( \frac{r}{2u - r} \right)^2$

Vedclass · Answer

(D) The mirror formula is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Since $f = \frac{r}{2}$,we have $\frac{1}{v} + \frac{1}{u} = \frac{2}{r}$.Differentiating both sides with respect to time $t$,we get $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$.Therefore,the velocity of the image $v_i = \frac{dv}{dt} = -\left( \frac{v}{u} \right)^2 \frac{du}{dt}$.Given the object is moving towards the mirror,$\frac{du}{dt} = -v_0$.From the mirror formula,$\frac{1}{v} = \frac{2}{r} - \frac{1}{u} = \frac{2u - r}{ru}$,which gives $v = \frac{ru}{2u - r}$.Substituting $v$ into the expression for $v_i$,we get $v_i = -\left( \frac{ru / (2u - r)}{u} \right)^2 (-v_0) = -v_0 \left( \frac{r}{2u - r} \right)^2$.

The speed at which the image of a luminous point object is moving,if the luminous point object is moving at speed $v_0$ towards a spherical mirror along its axis,is (Given: $r =$ radius of curvature,$u =$ object distance):

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