In an experiment for estimating the value of focal length of converging mirror, image of an object placed at $$40$$ cm from the pole of the mirror is formed at 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 $$\frac{1}{K}$$ cm. The value of $$K$$ is ______.

Solution :

For a converging mirror,

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

Given :

$$u = 40\text{ cm}$$

$$v = 120\text{ cm}$$

Therefore,

$$\frac{1}{f} = \frac{1}{120} + \frac{1}{40}$$

$$= \frac{1+3}{120}$$

$$= \frac{1}{30}$$

$$f = 30\text{ cm}$$

Modified scale has :

$$20\text{ divisions} = 1\text{ cm}$$

Least count :

$$LC = \frac{1}{20}\text{ cm}$$

$$= 0.05\text{ cm}$$

Hence,

$$\Delta u = \Delta v = 0.05\text{ cm}$$

Using error propagation formula :

$$\frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}$$

Substituting values :

$$\Delta f = (30)^2\left(\frac{0.05}{(120)^2} + \frac{0.05}{(40)^2}\right)$$

$$= 900 \times 0.05 \left(\frac{1}{14400} + \frac{1}{1600}\right)$$

$$= 45\left(\frac{1+9}{14400}\right)$$

$$= 45 \times \frac{10}{14400}$$

$$= \frac{450}{14400}$$

$$= \frac{1}{32}\text{ cm}$$

Given,

$$\Delta f = \frac{1}{K}\text{ cm}$$

Therefore,

$$K = 32$$

Final Answer :

$$32$$