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NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Units and Dimensions

NTA Abhyas 2020 · Physics · Units and Dimensions

The focal length of a mirror is given by (2)/(f) = (1)/(v) -...

The focal length of a mirror is given by

\dfrac{2}{f} = \dfrac{1}{v} - \dfrac{1}{u}

. If in the measurement of

u

and

v

, the errors are equal to

p

each, then the relative error in

f

is

Options

A $\dfrac{p}{2} \left(\dfrac{1}{u} + \dfrac{1}{v}\right)$
B $p \left(\dfrac{1}{u} + \dfrac{1}{v}\right)$
C $\dfrac{p}{2} \left(\dfrac{1}{u} - \dfrac{1}{v}\right)$
D $p \left(\dfrac{1}{u} - \dfrac{1}{v}\right)$

Correct answer

B. $p \left(\dfrac{1}{u} + \dfrac{1}{v}\right)$

Step-by-step solution

Given, equation is

\dfrac{2}{f} = \dfrac{1}{v} - \dfrac{1}{u}

....(i)<br /><br />Differentiating the given equation, we have<br /><br />

-

\dfrac{2}{f2}

df = -

\dfrac{1}{v2}

dv - \left(-

\dfrac{1}{u2}\right)

du

<br /><br />

= - p \left(

\dfrac{1}{v} - \dfrac{1}{u}

\right)\left(

\dfrac{1}{v} + \dfrac{1}{u}\right)\left(∵ dv = du = p\right)

=\dfrac{- 2p}{f}\left(

\dfrac{1}{v} + \dfrac{1}{u}\right)

& & & & & & & & [using Eq. (i)]

∴

\dfrac{df}{f} =

p \left(

\dfrac{1}{v} + \dfrac{1}{v}$$\right)

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