When a concave mirror is placed facing the sun, the sun rays converge to a point 15 cm from the mirror. Now, an erect 2 cm long pin is placed 30 cm away on the principal axis of the mirror. Where a card should be placed so that the image of the pin can be taken on the card? What would be the nature and height of the image?

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Solution

f = -15 cm.

ho = 2 cm

u = -30 cm

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

$\frac{1}{v}$ = $\frac{1}{- 15}$ + $\frac{1}{30}$

v = -30 cm

Image is real. The card must be placed 30 cm from the pole of the mirror on the same side of the object.

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

$\frac{h_{i}}{2}$ = $\frac{30}{- 30}$

$h_{i}$ = -2.

Image is inverted and its height is 2 cm

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f = -15 cm. ho = 2 cm u = -30 cm Mirror formula 1f = 1v + 1u 1v = 1−15 + 130 v = -30 cm Image is real. The card must be placed 30 cm from the pole of the mirror on the same side of the object. (hi)(ho) = −vu hi2 = 30−30 hi = -2. Image is inverted and its height is 2 cm