A rod of length 10 cm lies along the principle axis of a concave mirror of focal length 10 cm as shown in figure. The length of the image is ______ cm. [2026]

Question

Smart Achievers · Accepted Answer

(2)

$For A$

$u = - 20 cm, f = - 10 cm$

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $∴$ $v = \frac{u \times f}{u - f} = - 20 cm$

$For B$

$u = - 30 cm, f = - 10 cm$

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

$v = \frac{u f}{u - f} = - 15 cm$

$Length of image = 20 - 15 = - 5 cm$

Solution

Q.
A rod of length 10 cm lies along the principle axis of a concave mirror of focal length 10 cm as shown in figure. The length of the image is ______ cm. [2026]

1 2.5

2 5

3 7.5

4 7

Ans.
(2)

$For A$

$u = - 20 cm, f = - 10 cm$

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $∴$ $v = \frac{u \times f}{u - f} = - 20 cm$

$For B$

$u = - 30 cm, f = - 10 cm$

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

$v = \frac{u f}{u - f} = - 15 cm$

$Length of image = 20 - 15 = - 5 cm$

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