Ray Optics

JEE Main level, numerical answer, medium

33 of 295

Question 33 of 295

In an experiment for estimating the value of focal length of converging mirror, image of an object placed at 40cm40cm from the pole of the mirror is formed at distance 120cm120cm from the pole of the mirror. These distances are measured with a modified scale in which there are 2020 small divisions in 1cm1cm. The value of error in measurement of focal length of the mirror is 1Kcm\dfrac{1}{K}cm. The value of KK is ______.

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Answer: 32

Given here, u=−40cm,v=−120cmu=-40cm,v=-120cm
Using mirror formula, 1v+1u=1f\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}
−1120−140=1f⇒f=−30cm\dfrac{-1}{120}-\dfrac{1}{40}=\dfrac{1}{f} \Rightarrow f=-30cm
Now, differentiating mirror equation, we get
−1v2dv−1u2du=−1f2df\dfrac{-1}{{v}^{2}}dv-\dfrac{1}{{u}^{2}}du=-\dfrac{1}{{f}^{2}}df
Also, the least count of scale, dv=du=120cmdv=du=\dfrac{1}{20}cm
Putting in above equation,
∴120(120)2+120(40)2=df(30)2∴\dfrac{\dfrac{1}{20}}{(120{)}^{2}}+\dfrac{\dfrac{1}{20}}{(40{)}^{2}}=\dfrac{df}{(30{)}^{2}}
On solving, we get
df=132cmdf=\dfrac{1}{32}cm
∴K=32∴K=32.

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