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 (integer)
32
Solution
<p>${1 \over v} + {1 \over u} = {1 \over f}$ ...... (1)</p>
<p>$$ \Rightarrow - {1 \over {{f^2}}}df = - {1 \over {{v^2}}}dv - {1 \over {{u^2}}}du$$</p>
<p>$\Rightarrow {{df} \over {{f^2}}} = {{dv} \over {{v^2}}} + {{du} \over {{u^2}}}$ ..... (2)</p>
<p>From (1) : $- {1 \over {120}} - {1 \over {40}} = {1 \over f} \Rightarrow f = - 30$ cm</p>
<p>Also, least count $= {{1\,\mathrm{cm}} \over {20}} = 0.05$ cm</p>
<p>$$ \Rightarrow df = \left[ {{{0.05} \over {{{120}^2}}} + {{0.05} \over {{{40}^2}}}} \right] \times {30^2}$$</p>
<p>$$ = 0.05\left[ {{1 \over {16}} + {9 \over {16}}} \right] = {5 \over 8} \times {5 \over {100}} = {1 \over {32}}$$ cm</p>
<p>$\Rightarrow k = 32$</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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