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Ray Optics

JEE Main level, single correct, medium

171 of 295

Question 171 of 295

A $2$ meter long scale with least count of $0.2cm$ is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at $80cm$ mark and $1m$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $180cm$ mark. The $%$ error in the estimation of focal length is:

Show answer and solution

Answer: (B) $1.70$

The formula for a lens is given by

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}...(i)

u=\left(100\pm 0.2\right)cm-\left(80\pm 0.2\right)cm=\left(20\pm 0.4\right)cm

v=\left(180\pm 0.2\right)cm-\left(100\pm 0.2\right)cm=\left(80\pm 0.4\right)cm

Substituting the values in the equation of focal length,

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{80}-\left(-\dfrac{1}{20}\right)=\dfrac{5}{80}

\Rightarrow f=16cm

Differentiating the equation (i)

\dfrac{df}{{f}^{2}}=\dfrac{dv}{{v}^{2}}+\dfrac{du}{{u}^{2}}

(Only positive sign is considered for error calculation)

=\dfrac{0.4}{6400}+\dfrac{0.4}{400}=0.4\times \dfrac{17}{6400}

\Rightarrow df=0.4\times \dfrac{17}{6400}\times {16}^{2}=0.272

The percentage error is

\dfrac{df}{f}\times 100=\dfrac{0.272}{16}\times 100=1.70%

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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