JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 21)
In an experiment for estimating the value of focal length of converging mirror, image of an object placed at $$40 \mathrm{~cm}$$ from the pole of the mirror is formed at distance $$120 \mathrm{~cm}$$ from the pole of the mirror. These distances are measured with a modified scale in which there are 20 small divisions in
$$1 \mathrm{~cm}$$. The value of error in measurement of focal length of the mirror is $$\frac{1}{\mathrm{~K}} \mathrm{~cm}$$. The value of $$\mathrm{K}$$ is __________.
Answer
32
Explanation
$${1 \over v} + {1 \over u} = {1 \over f}$$ ...... (1)
$$ \Rightarrow - {1 \over {{f^2}}}df = - {1 \over {{v^2}}}dv - {1 \over {{u^2}}}du$$
$$ \Rightarrow {{df} \over {{f^2}}} = {{dv} \over {{v^2}}} + {{du} \over {{u^2}}}$$ ..... (2)
From (1) : $$ - {1 \over {120}} - {1 \over {40}} = {1 \over f} \Rightarrow f = - 30$$ cm
Also, least count $$ = {{1\,\mathrm{cm}} \over {20}} = 0.05$$ cm
$$ \Rightarrow df = \left[ {{{0.05} \over {{{120}^2}}} + {{0.05} \over {{{40}^2}}}} \right] \times {30^2}$$
$$ = 0.05\left[ {{1 \over {16}} + {9 \over {16}}} \right] = {5 \over 8} \times {5 \over {100}} = {1 \over {32}}$$ cm
$$ \Rightarrow k = 32$$
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