QJEEAdv23Ph35a8: JEE Advanced 2023 Physics - Class 11 JEEAdv23Ph35a8 Solution: In an experiment for determination of the focal length of a…

Given :

- Object distance $u = 10.0 \pm 0.1 \, \text{cm}$

- Image distance $v = 20.0 \pm 0.2 \, \text{cm}$

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 :

$f = \frac{1}{\left(\frac{1}{v} - \frac{1}{u}\right)} = \frac{1}{\left(\frac{1}{20 \, \text{cm}} - \frac{1}{-10 \, \text{cm}}\right)} = \frac{1}{0.05 \, \text{cm}^{-1} + 0.1 \, \text{cm}^{-1}} = \frac{1}{0.15 \, \text{cm}^{-1}} = \frac{20}{3} \, \text{cm}$

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 :

$\frac{df}{f} \times 100 = \left(\frac{1}{v^2} dv + \frac{1}{u^2} du\right) \times \frac{f}{1} \times 100$

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:

$\frac{df}{f} \times 100 = \left(\frac{0.2}{20^2} + \frac{0.1}{10^2}\right) \times \frac{20}{3} \times 100 = 1\%$

So, the error in the focal length of the lens is indeed 1% (i.e., $n = 1$ ).