Ray Optics Question 9
Question 9 - 30 January - Shift 1
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 $1 / K$ $cm$. The value of $K$ is
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Answer: (32)
Solution:
Formula: Mirror Formula
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{-1}{120}-\frac{1}{40}=\frac{1}{f}, \quad f=-30 cm$
Now,
$\frac{-1}{v^{2}} d v-\frac{1}{u^{2}} d u=-\frac{1}{f^{2}} d f$
Also $dv=du=\frac{1}{20} cm$
$\frac{\frac{1}{20}}{(120)^{2}}+\frac{\frac{1}{20}}{(40)^{2}}=\frac{d f}{(30)^{2}}$
On solving
$ \begin{aligned} & df=\frac{1}{32} cm \\ & \therefore \quad k=32 \end{aligned} $
