A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 20 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image ?

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Solution

Given that, $f_{1} = 15 cm$ , $F_{m} = 10 cm$ , $h_{0} = 2 cm$

The object is placed 30 cm from lens.

From lens formula,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow v = \frac{u f}{u + f}$

Since, u=-30 cm and f = 15 cm

So, $v = \frac{(- 30) \times 15}{(- 30 + 15)}$

$= \frac{- 450}{- 15} = 30 cm$

So, real and inverted image (AB) will be formed at 30 cm from the lens and it will be of same size as the object. Now this real image is at a distance 20 cm from the concave mirror. Since, $f_{m} = 10$ , this real image is at the centre of curvature of the mirror. So the mirror will form an inverted image $A^{''} B^{''}$ at the same place of same size. Again, due to refraction in the lens the final image will be formed at AB and will be of same size as that of object. $(A^{''} B^{''})$

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