In an experiment for estimating the value of focal length of a converging mirror, the image of an object placed at 40 cm from the pole of the mirror is formed at a distance 120 cm from the pole of the mirror. [2023]

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Solution

Q.
In an experiment for estimating the value of focal length of a converging mirror, the image of an object placed at 40 cm from the pole of the mirror is formed at a distance 120 cm from the pole of the mirror. These distances are measured with a modified scale in which there are 20 small divisions in 1 cm. The value of error in measurement of focal length of the mirror is 1/K cm. The value of K is _____. [2023]

Ans.
(32)

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

$\frac{- 1}{120} - \frac{1}{40} = \frac{1}{f}, f = - 30 cm$

$Now, \frac{- 1}{v^{2}} d v - \frac{1}{u^{2}} d u = - \frac{1}{f^{2}} d f$

$Also d v = d u = \frac{1}{20} cm$

$∴$ $\frac{\frac{1}{20}}{{(120)}^{2}} + \frac{\frac{1}{20}}{{(40)}^{2}} = \frac{d f}{{(30)}^{2}}$

$On solving d f = \frac{1}{32} cm \Rightarrow k = 32$

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