A converging lens of focal length 15cm and a converging mirror of focal length 10cm are placed 50cm apart. If a pin of length 2.0cm is placed 30cm 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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Given that, $f_1 = 15cm, F_m = 10cm, h_o = 2cm$
The object is placed 30cm from lens $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}.$
$\Rightarrow\text{v}=\frac{\text{uf}}{\text{u}+\text{f}}$
Since, u = -30cm and f = 15cm
So, v = 30cm
So, real and inverted image (A'B') will be formed at 30cm from the lens and it will be of same size as the object. Now, this real image is at a distance 20cm from the concave mirror. Since, $f_m = 10cm$, 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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