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QJEEM23Ph35ec: JEE Main 2023 Physics - Class 12 JEEM23Ph35ec Solution: In an experiment for estimating the value of focal length of…

In an experiment for estimating the value of focal length of converging mirror, image of an object placed at 40 cm40 \mathrm{~cm} from the pole of the mirror is formed at distance 120 cm120 \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 cm1 \mathrm{~cm} . The value of error in measurement of focal length of the mirror is 1 K cm\frac{1}{\mathrm{~K}} \mathrm{~cm} . The value of K\mathrm{K} is __________.

Step-by-Step Verified Solution

1v+1u=1f{1 \over v} + {1 \over u} = {1 \over f} ...... (1)

⇒−1f2df=−1v2dv−1u2du\Rightarrow - {1 \over {{f^2}}}df = - {1 \over {{v^2}}}dv - {1 \over {{u^2}}}du

⇒dff2=dvv2+duu2\Rightarrow {{df} \over {{f^2}}} = {{dv} \over {{v^2}}} + {{du} \over {{u^2}}} ..... (2)

From (1) : −1120−140=1f⇒f=−30- {1 \over {120}} - {1 \over {40}} = {1 \over f} \Rightarrow f = - 30 cm

Also, least count =1 cm20=0.05= {{1\,\mathrm{cm}} \over {20}} = 0.05 cm

⇒df=[0.051202+0.05402]×302\Rightarrow df = \left[ {{{0.05} \over {{{120}^2}}} + {{0.05} \over {{{40}^2}}}} \right] \times {30^2}

=0.05[116+916]=58×5100=132= 0.05\left[ {{1 \over {16}} + {9 \over {16}}} \right] = {5 \over 8} \times {5 \over {100}} = {1 \over {32}} cm

⇒k=32\Rightarrow k = 32

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