A car is fitted with a convex side-view mirror of focal length 20cm. A second car 2.8m behind the first car is overtaking the first car at relative speed of 15m/s. The speed of the image of the second car as seen in the mirror of the first one is

Given data: focal length 20 cm, relative speed $15 \mathrm{m}/\mathrm{s}$.

Concept used: combination of lenses 

Detailed solution:

We know that, Mirror formula is $\frac{1}{\vartheta }+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{\vartheta }=\frac{1}{280}=\frac{1}{20}$

$\Rightarrow \frac{1}{\vartheta }=\frac{1}{20}+\frac{1}{280}=\frac{15}{280}$

Upon differentiation 

$\frac{-1}{{\vartheta }^2}\frac{\mathrm{d}\vartheta }{\mathrm{dt}}-\frac{1}{{\mathrm{u}}^2}\frac{\mathrm{du}}{\mathrm{dt}}=0$

$-\frac{15}{280}\times \frac{15}{280}{\vartheta }_1=\frac{1}{280}\times \frac{1}{280}\times 15$

${\vartheta }_{\mathrm{i}}=\frac{\mathrm{du}}{\mathrm{dt}}=\frac{1}{15}{ \mathrm{ms}}^{-1}$

Hence, option (A) is the solution.