Question

Magnitude of velocity of image at the given instant

• A. $0.08cm{s}^{-1}$
• B. $0.64cm{s}^{-1}$
• C. $0.16cm{s}^{-1}$
• D. $0.32cm{s}^{-1}$

Answer 1

The correct option is C $0.16cm{s}^{-1}$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ -(1)

Differentiating w. r.t to t, $-\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}=0$

$⟹\frac{dv}{dt}=-\frac{v^2}{u^2}\frac{du}{dt}$ -(2)

where $\frac{du}{dt}and\frac{dv}{dt}$ represents the velocity of object and its image respectively.

Given: $u=-30cmf=20cm\frac{du}{dt}=1cm{s}^{-1}$

From Equation (1), $\frac{1}{v}+\frac{1}{-30}=\frac{1}{20}$

$⟹v=12cm$

From (2), $\frac{dv}{dt}=-\frac{{12}^2}{(-30{)}^2}\times 1$

$\frac{dv}{dt}=-0.16cm{s}^{-1}$ means image moves towads the pole.

$|\frac{dv}{dt}|=0.16cm{s}^{-1}$