Question

6. An object 5 cm high is placed at a distance 60 cm
in front of a concave mirror of focal length 10 cm.
Find (i) the position and (ii) size, of the image.
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Answer 1

$\large\sf{\underline{\underline{Given}}}$

$\sf Size\:of\:object,\:h_o = 5cm$

$\sf Size\:of\:the\:image,\:h_i = ?$

$\sf Focal\: length,\:f = -10cm$

$\sf Object\:distance,\:u = -60cm$

$\sf Image\: position,\:v = ?$

$\large\sf{\underline{\underline{Solution}}}$

$\sf Using\:mirror\:formula,$

$\large\sf \frac{1}{f} = \frac{1}{v}+\frac{1}{u}$

$\large\sf \frac{1}{ - 10} = \frac{1}{v} + \frac{1}{ -60}$

$\large\sf \frac{1}{v} = \frac{ - 6 + 1}{60}$

$\sf v = -12cm$

(i) Hence, the position of the image is at 12cm in front of the mirror.

$\sf Magnification\:(M)=$ $\large\sf \frac{h_i}{h_o} = \frac{ - v}{u} = \frac{12}{ - 60} = - 0.2$

$\sf h_i = -0.2 × h_o$

$\sf h_i = -0.2 × 5 = -1cm$

(ii) Hence, the size of inverted image is 1cm