First slide

Spherical/Curved mirrors

Question

A small piece of wire bent into an L shape, with upright and horizontal portions of equal lengths, is placed with the horizontal portion along the axis of the concave mirror whose radius of curvature is 10 cm. If the bend is 20 cm from the pole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portions of the wire is

Moderate

A

1 : 2
B

3 : 1
C

1 : 3
D

2 : 1

Solution

For concave mirror, focal length = $\frac{R}{2} = \frac{10}{2}$ For upright portion, Transverse Magnification $m = \frac{f}{f - u} = \frac{\frac{- 10}{2}}{\frac{- 10}{2} - (- 20)} = \frac{- 5}{- 5 + 20} = \frac{- 5}{15} = - \frac{1}{3}$
For horizontal portion, for small size objects, Longitudinal magnification is $= {(\frac{v}{u})}^{2} = {(\frac{1}{3})}^{2}$
i.e., $\frac{1}{9}$
Required ratio is $\frac{- 1 / 3}{1 / 9} = - 3 : 1$

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