An object and a concave mirror of focal length both move along the principal axis of the mirror with constant speeds. The object moves with speed towards the mirror with respect to the laboratory frame. The distance between the object and the mirror at

Question

An object and a concave mirror of focal length $f = 10 cm$ both move along the principal axis of the mirror with constant speeds. The object moves with speed $V_{0} = 15 cm s^{- 1}$ towards the mirror with respect to the laboratory frame. The distance between the object and the mirror at a given moment is denoted by $u$ . When $u = 30 cm$ , the speed of the mirror $V_{m}$ is such that the image is instantaneous at rest with respect to the laboratory frame, and the object forms a real image. The magnitude of $V_{m}$ is _____ $cm s^{- 1}$ . [2022]

Smart Achievers · Accepted Answer

(3)

$We have, u = - 30 cm, f = - 10 cm$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} - \frac{1}{- 30} = \frac{1}{30} - \frac{1}{10} = \frac{1 - 3}{30} = - \frac{1}{15}$

So, $v = - 15 cm$

As, ${\vec{V}}_{I, M} = - {(\frac{v}{u})}^{2} {\vec{V}}_{O, M}$

$\Rightarrow {\vec{V}}_{I} - {\vec{V}}_{M} = - {(\frac{v}{u})}^{2} ({\vec{V}}_{O} - {\vec{V}}_{M})$

$\Rightarrow 0 - {\vec{V}}_{M} = - \frac{1}{4} (15 \hat{i} - {\vec{V}}_{M})$

$\Rightarrow - {\vec{V}}_{M} = - \frac{15}{4} \hat{i} + \frac{1}{4} {\vec{V}}_{M}$

$\Rightarrow - \frac{5}{4} {\vec{V}}_{M} = - \frac{15}{4} \hat{i}$

$∴$ ${\vec{V}}_{M} = 3 \hat{i}$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (3) We have, u=-30 cm, f=-10 cm 1v=1f-1u=1-10-1-30=130-110=1-330=-115 So, v=-15 cm As, V→I,M=-(vu)2V→O,M ⇒ V→I-V→M=-(vu)2(V→O-V→M) ⇒ 0-V→M=-14(15i^-V→M) ⇒ -V→M=-154i^+14V→M ⇒ -54V→M=-154i^ ∴ V→M=3i^

Want Us to Email you About Special Offers & Updates?