Question 301: Units , Dimensions and Error Analysis

Question

In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is $10 \pm 0.1 \mathrm{~cm}$ and the distance of its real image from the lens is $20 \pm 0.2 \mathrm{~cm}$. The error in the determination of focal length of the lens is $n \%$. The value of $n$ is ______.

Accepted Answer

Given :
- Object distance $u = 10.0 \pm 0.1 \, 	ext{cm}$
- Image distance $v = 20.0 \pm 0.2 \, 	ext{cm}$
According to the lens formula for a thin lens :
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
So, we can calculate the focal length :
$
f = \frac{1}{\left(\frac{1}{v} - \frac{1}{u}ight)} = \frac{1}{\left(\frac{1}{20 \, 	ext{cm}} - \frac{1}{-10 \, 	ext{cm}}ight)} = \frac{1}{0.05 \, 	ext{cm}^{-1} + 0.1 \, 	ext{cm}^{-1}} = \frac{1}{0.15 \, 	ext{cm}^{-1}} = \frac{20}{3} \, 	ext{cm}
$
Next, we need to calculate the error in the determination of the focal length. For that, we find the differential of the lens formula :
Then, the derivative of the equation gives us the change in the focal length $(df)$ in terms of the changes in the object distance $(du)$ and the image distance $(dv)$:
$-\frac{1}{v^2} dv + \frac{1}{u^2} du = \frac{-1}{f^2} df$
For maximum error, we get :
$\frac{1}{f^2} df = \frac{1}{v^2} dv + \frac{1}{u^2} du$
This equation tells us how errors in $(u)$ and $v$ propagate to an error in $f$. Now, when you compute the relative error in the focal length, you get :
$\frac{df}{f} 	imes 100 = \left(\frac{1}{v^2} dv + \frac{1}{u^2} duight) 	imes \frac{f}{1} 	imes 100$
Plugging in your values of $u = 10 \mathrm{~cm}$, $du = 0.1 \mathrm{~cm}$, $v = 20 \mathrm{~cm}$, $dv = 0.2 \mathrm{~cm}$, and $f = 20/3 \mathrm{~cm}$, you indeed get:
$\frac{df}{f} 	imes 100 = \left(\frac{0.2}{20^2} + \frac{0.1}{10^2}ight) 	imes \frac{20}{3} 	imes 100 = 1\%$
So, the error in the focal length of the lens is indeed 1% (i.e., $n = 1$).

Units , Dimensions and Error Analysis

Integer / Numeric Answer Type