Question

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 cm and the distance of its real image from the lens is 20 ± 0.2 cm. The error in the determination of focal length of the lens is 𝑛 %. The value of 𝑛 is ____ .

Answer 1

Object distance \(= 10 ± 0.1\) cm Image distance \(= 20 ± 0.2\) cm 
Applying lens formula ….. (1)

\(\frac {1}{20} - \frac{1}{-10} = \frac {1}{f}\)

f = \(\frac{20}{3} cm\)

differentiate equation 1: 

\(\frac {1}{f^2}df = +\frac{1}{v^2}dv+\frac{1}{u^2}du\)

\(\frac{df}{f}\times 100 = \frac{0.2}{20^2}+\frac{0.1}{10^2}\times \frac{20}{3}\times 100\)

= 1 

\(\frac{df}{f}\times 100 = 1%\)

So, the answer is \(1\).

Answer 2

Given,
Object distance = 10 ± 0.1 cm
Image distance = 20 ± 0.2 cm
Applying lens formula
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
Now put the value in this equation
\(\frac{1}{20}-\frac{1}{(-10)}=\frac{1}{f}\)
\(\frac{3}{20}=\frac{1}{f}\)
\(f=\frac{20}{3}cm\)

Now Differentiate the equation
\(-\frac{1}{v^2}dv+\frac{1}{u^2}du=\frac{-1}{f^2}df\)

\(\frac{1}{f^2}df=\frac{1}{v^2}dv+\frac{-1}{u^2}du\)

\((\frac{df}{f})\times 100=(\frac{0.2}{20^2}+\frac{0.1}{10^2})\frac{20}{3}\times 100\)

\((\frac{0.2}{20^2}+\frac{0.1}{10^2})\frac{20}{3}=1\)

\(\frac{df}{f}\times 100=1\%\)

So, the answer is 1.