A 1 cm high object is placed at a distance of 32 cm from a concave mirror. The image is real, inverted and 1.5 cm in size. Calculate the focal length of the mirror and the position of the image.

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Solution

u = -32 cm

$h_{o}$ = 1cm

$h_{i}$ = -1.5

$\frac{h_{i}}{h_{o}}$ = $\frac{- v}{u}$

$\frac{- 1.5}{1}$ = $\frac{- v}{- 32}$

v = -48 cm

Image is 48 cm from the pole and on the same side of the object

Mirror formula $\frac{1}{f}$ = $\frac{1}{v}$ + $\frac{1}{u}$

$\frac{1}{f}$ = $\frac{1}{f}$ - $\frac{1}{32}$

f = -19.2 cm

Focal length of the mirror is 19.2 cm

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A 1 cm high object is placed at a distance of 32 cm from a concave mirror. The image is real, inverted and 1.5 cm in size. Calculate the focal length of the mirror and the position of the image.

u = -32 cm ho = 1cm hi = -1.5 hiho = −vu −1.51 = −v−32 v = -48 cm Image is 48 cm from the pole and on the same side of the object Mirror formula 1f = 1v + 1u 1f =1f - 132 f = -19.2 cm Focal length of the mirror is 19.2 cm