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  • 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 mirr

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

Option: 1

32


Option: 2

__


Option: 3

__


Option: 4

__


Answers (1)

best_answer

\begin{aligned} & \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ & \mathrm{f}^{-1}=v^{-1}+u^{-1} \end{aligned}             \mathrm{dv}=\mathrm{du}=\frac{1 \mathrm{~cm}}{20}=0.05 \mathrm{~cm} \text { (given) }

\begin{aligned} & (-1) f^{-2} d f=(-1) v^{-2} d v-u^{-2} d u \\ & \frac{d f}{f^2}=\frac{d v}{v^2}+\frac{d u}{u^2} \\ & \frac{1}{f}=\frac{1}{(-120)}+\frac{1}{-40} \\ & \frac{1}{f}=\frac{1+3}{(-120)}=\frac{4}{-120} \Rightarrow(1) \end{aligned}__________(i)

Put value of f, du, dv in (1)

\begin{aligned} & \frac{\mathrm{df}}{(30)^2}=\frac{0.05}{(120)^2}+\frac{0.05}{(40)^2} \\ & \text { df }=\frac{1}{32} \mathrm{~cm} \quad \text { so } \mathrm{K}=32 \\ & \end{aligned}

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Ajit Kumar Dubey

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