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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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):

Similar Questions

There are two plane mirrors with reflecting surfaces facing each other. The mirrors are moving with speed $v$ away from each other. $A$ point object is placed between the mirrors. The velocity of the $n$-th image will be

Choose the correct statement$(s)$ related to the motion of an object and its image in the case of mirrors.

Find the velocity of the image in the situation shown in the figure. (All velocities are in $m/s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module