Focal length of the plano-convex lens is 15 cm. A small object is placed at A as shown in the figure. The plane surface is silvered. The image will form at [2006]

Question

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(2)

The focal length $f$ of the equivalent mirror when lens is silvered,

$\frac{1}{f} = \frac{2}{f_{l}} + \frac{1}{f_{m}} = \frac{2}{15} + \frac{1}{\infty} \Rightarrow f = \frac{15}{2} cm$

Since $f$ has a positive value, the combination behaves as a converging mirror.

Here, $u = - 20 cm, f = - \frac{15}{2} cm, v = ?$

According to the mirror formula, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{v} - \frac{1}{- 20} = \frac{1}{- \frac{15}{2}}$ $\Rightarrow v = - 12 cm$

Negative sign indicates that the image is 12 cm in front or left of the mirror.

Solution

Q.
Focal length of the plano-convex lens is 15 cm. A small object is placed at A as shown in the figure. The plane surface is silvered. The image will form at [2006]

1 60 cm to the left of lens

2 12 cm to the left of lens

3 60 cm to the right of lens

4 30 cm to the left of lens

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