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Units and Measurements question

2023 · Shift 1 · Q44
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Units and Measurements question

2023 · Shift 1 · Q44

JEE AdvancedPhysicsUnits and MeasurementsNumerical+4 / −1
In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is 10±0.1 cm10 \pm 0.1 \mathrm{~cm}10±0.1 cm and the distance of its real image from the lens is 20±0.2 cm20 \pm 0.2 \mathrm{~cm}20±0.2 cm. The error in the determination of focal length of the lens is n%n \%n%. The value of nnn is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 1

  1. For a thin convex lens, the focal length fff is related to object distance uuu and image distance vvv by
1f=1u+1v\frac{1}{f}=\frac{1}{u}+\frac{1}{v}f1​=u1​+v1​

So,

f=uvu+vf=\frac{uv}{u+v}f=u+vuv​

Given:

u=10±0.1 cm,v=20±0.2 cmu=10\pm 0.1\text{ cm}, \qquad v=20\pm 0.2\text{ cm}u=10±0.1 cm,v=20±0.2 cm
  1. First calculate the focal length:
f=10×2010+20=20030=203 cmf=\frac{10\times 20}{10+20}=\frac{200}{30}=\frac{20}{3}\text{ cm}f=10+2010×20​=30200​=320​ cm
  1. Now calculate the maximum fractional error. Since
f=uvu+vf=\frac{uv}{u+v}f=u+vuv​

take logarithmic error:

Δff=Δuu+Δvv+Δ(u+v)u+v\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}+\frac{\Delta (u+v)}{u+v}fΔf​=uΔu​+vΔv​+u+vΔ(u+v)​

But since (u+v)(u+v)(u+v) is in the denominator, its contribution is subtracted in differentiation, and for maximum error we take magnitude:

Δff=Δuu+Δvv+Δ(u+v)u+v\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}+\frac{\Delta (u+v)}{u+v}fΔf​=uΔu​+vΔv​+u+vΔ(u+v)​

with

Δ(u+v)=Δu+Δv=0.1+0.2=0.3\Delta (u+v)=\Delta u+\Delta v=0.1+0.2=0.3Δ(u+v)=Δu+Δv=0.1+0.2=0.3

A cleaner way is to use:

f=uvu+vf=\frac{uv}{u+v}f=u+vuv​

so

Δff=vu+vΔuu+uu+vΔvv\frac{\Delta f}{f}=\frac{v}{u+v}\frac{\Delta u}{u}+\frac{u}{u+v}\frac{\Delta v}{v}fΔf​=u+vv​uΔu​+u+vu​vΔv​

But for JEE error propagation through reciprocals is easiest:

From

1f=1u+1v\frac{1}{f}=\frac{1}{u}+\frac{1}{v}f1​=u1​+v1​

Let F=1fF=\frac{1}{f}F=f1​. Then

ΔF=Δ(1u)+Δ(1v)\Delta F=\Delta\left(\frac{1}{u}\right)+\Delta\left(\frac{1}{v}\right)ΔF=Δ(u1​)+Δ(v1​)

Using

Δ(1u)=Δuu2,Δ(1v)=Δvv2\Delta\left(\frac{1}{u}\right)=\frac{\Delta u}{u^2}, \qquad \Delta\left(\frac{1}{v}\right)=\frac{\Delta v}{v^2}Δ(u1​)=u2Δu​,Δ(v1​)=v2Δv​

So,

Δ(1f)=0.1102+0.2202=0.1100+0.2400=0.001+0.0005=0.0015\Delta\left(\frac{1}{f}\right)=\frac{0.1}{10^2}+\frac{0.2}{20^2} =\frac{0.1}{100}+\frac{0.2}{400} =0.001+0.0005=0.0015Δ(f1​)=1020.1​+2020.2​=1000.1​+4000.2​=0.001+0.0005=0.0015

Also,

1f=110+120=0.1+0.05=0.15\frac{1}{f}=\frac{1}{10}+\frac{1}{20}=0.1+0.05=0.15f1​=101​+201​=0.1+0.05=0.15

Hence fractional error in 1f\frac{1}{f}f1​ is

Δ(1/f)1/f=0.00150.15=0.01\frac{\Delta (1/f)}{1/f}=\frac{0.0015}{0.15}=0.011/fΔ(1/f)​=0.150.0015​=0.01

Since fractional error in fff is same as fractional error in 1/f1/f1/f,

Δff=0.01\frac{\Delta f}{f}=0.01fΔf​=0.01

Therefore percentage error is

0.01×100=1%0.01\times 100 = 1\%0.01×100=1%
  1. Hence,
n=1n=1n=1
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