First slide

Spherical/Curved mirrors

Question

A cube of side length 1 mm is placed at distance 30 cm from the pole of a concave mirror of focal length 20 cm. Its one side is parallel optical axis as shown in the figure. Find the magnification in the volume of cube

Moderate

A

4
B

2
C

16
D

24

Solution

Here, u=-30cm; f= -20cm;
$∵ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{30} = - \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{30} - \frac{1}{20} = \frac{2 - 3}{60} ∴ v = - 60 c m$
$∴$ Transverse magnification is $m = - \frac{v}{u} = - 2$
And longitudinal magnification is $m_{1} = \frac{- v^{2}}{u^{2}} = - 4$
The magnification in sides parallel to optical axis is m1 and the magnification in sides perpendicular to optical axis is m.
$∴ \frac{l'}{l} = \frac{- v^{2}}{u^{2}}$
And $l' = - 4 l = - 4 m m$
And $\frac{b'}{b} = m = - 2$
$b' = - 2 \times 1 = - 2 m m$
$∴$ Volume of image = $l' \times b' \times b'$
$= 4 \times (2) (2) = 16 m m^{3}$
Thus, the magnification in the volume is 16.

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