# Samarth plans to go from Delhi to Gujrat by a car. He thought of getting the car service done before going because the mirrors of his car were broken. After the car came from the agency he looked at all the mirrors and observed that a convex mirror used for rear-view has a radius of curvature of 2.00 m and the bus is located 4.00 m from the mirror. (i) What is the focal length of the mirror? (a) 0.5 m (b) 1 m (c) 2 m (d) 4 m (ii) What will be the image distance? (a) 0.2 m (b) 0.4 m (c) 0.8 m (d) 0.9 m (iii) What is the magnification in the given case? (a) 0.1 (b) 0.2 (c) 0.4 (d) 0.8 (iv) What is the nature of the image formed? (a) Virtual, erect and diminished (b) Real, inverted and diminished (c) Virtual, erect and magnified (d) Real, inverted and magnified

(i) Focal length of a lens, f = radius of curvature of the lens / 2

So,

$f=\frac{R}{2}=\frac{2.00m}{2}=1m\phantom{\rule{0ex}{0ex}}Ans\left(b\right)\phantom{\rule{0ex}{0ex}}$

(ii) Let v be the image distance.

Object distance, u = 4.00 m

Using the lens equation:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}U\mathrm{sin}gsignconventions:\phantom{\rule{0ex}{0ex}}\frac{1}{v}+\frac{1}{-4}=\frac{1}{1}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=1+\frac{1}{4}=\frac{5}{4}\phantom{\rule{0ex}{0ex}}Thus,\phantom{\rule{0ex}{0ex}}v=\frac{4}{5}m=0.8m\phantom{\rule{0ex}{0ex}}Ans\left(c\right)$

(iii) Magnification, M, is defined as:

$M=\frac{imageheight}{objectheight}=-\frac{v}{u}=-\frac{0.8}{-4}=+\frac{8}{40}=\frac{2}{5}=0.4\phantom{\rule{0ex}{0ex}}Ans\left(c\right)$

(iv) Since, the magnification is positive, thus the image formed is virtual, erect and diminished. Ans (a)

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