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} $