Question

A convex mirror used for rear view on a automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00m from this mirror.find the position,nature and size of image

Answer 1

$\large{\underline{\bf{\purple{Given:-}}}}$

✦ Radius of curvature ,R = +3.00m

✦ Object-distance ,u = -5.00m

$\large{\underline{\bf{\purple{To\:Find:-}}}}$

✦ we need to find the position , nature and size of the image.

$\huge{\underline{\bf{\red{Solution:-}}}}$

Focal length, F = R/2

$\leadsto \rm\:\:\frac{+3.00m}{3}$

$\leadsto \rm\:\:+1.50m$

[ principal focus of a convex mirror is behind the mirror.]

$\leadsto \bf\:\: \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \\ \leadsto \bf\:\: \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ \\\leadsto \rm\:\: + \frac{1}{1.50} - \frac{1}{( - 5.00)} \\ \\\leadsto \rm\:\: \frac{1}{1.50} + \frac{1}{5.00} \\ \\ \leadsto \rm\:\: \frac{5.00 + 1.50}{7.50} \\ \\ \leadsto \rm\:\:v = + \frac{7.50}{6.50} \\ \\ \leadsto \rm\:\: + 1.15m$

The image is 1.15m at the back of the mirror.

Magnification, m = h'/h = -v/u

$\leadsto \rm\:\:-\frac{1.15m}{-5.00m}$

$\leadsto \bf\:\:+0.23$

✦✦The image is virtual , erect and smaller in size by a factor of 0.23.

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