In an experiment to measure focal length (f) of convex lens, the least counts of the measuring scales for the position of object (u) and for the position of image (v) are ∆u and ∆v, respectively. The error in the measurement of the focal length of the convex lens will be :

The correct answer is f²[Δu/u² + Δv/v²].

To find the error in the focal length, we start from the standard lens formula and apply the rules of error propagation carefully.

Step 1: Start with the Lens Formula

The thin lens formula relating focal length f, object distance u, and image distance v is:

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

Here, u is negative (object on the left) by the sign convention, but for the purpose of error analysis we track the magnitudes of the uncertainties. The typical form used in error analysis (treating u and v as positive magnitudes for a real image) is:

\frac{1}{f}=\frac{1}{v}+\frac{1}{u}

This is because in a typical lab setup, u is measured as a positive magnitude (object distance), and for a convex lens forming a real image, the signed lens formula \frac{1}{f}=\frac{1}{v}-\frac{1}{u} with u negative gives the same form. For error propagation, what matters is that f is a function of both u and v.

Step 2: Rewrite in a Differentiable Form

Let us define:

F=\frac{1}{f}=\frac{1}{v}+\frac{1}{u}

Step 3: Apply Error Propagation to the Reciprocal Form

Since \frac{1}{f} is a sum of two independent terms, the absolute error in \frac{1}{f} is the sum of absolute errors in each term (errors always add for independent measurements):

\Delta \left(\frac{1}{f}\right)=\Delta \left(\frac{1}{v}\right)+\Delta \left(\frac{1}{u}\right)

Step 4: Find the Error in Each Reciprocal Term

For any quantity x, the error in \frac{1}{x} is found by differentiation:

d\left(\frac{1}{x}\right)=-\frac{dx}{x^2}

Taking the magnitude (absolute value) for error analysis:

\Delta \left(\frac{1}{v}\right)=\frac{\Delta v}{v^2},\Delta \left(\frac{1}{u}\right)=\frac{\Delta u}{u^2}

Therefore:

\Delta \left(\frac{1}{f}\right)=\frac{\Delta u}{u^2}+\frac{\Delta v}{v^2}

Step 5: Convert to Error in f Itself

Now, we need to relate \Delta \left(\frac{1}{f}\right) to \Delta f.

Differentiating \frac{1}{f} with respect to f:

d\left(\frac{1}{f}\right)=-\frac{df}{f^2}

Taking the magnitude:

\Delta \left(\frac{1}{f}\right)=\frac{\Delta f}{f^2}

Therefore:

\frac{\Delta f}{f^2}=\frac{\Delta u}{u^2}+\frac{\Delta v}{v^2}

Step 6: Final Result

Multiplying both sides by f²:

\Delta f=f^2[\frac{\Delta u}{u^2}+\frac{\Delta v}{v^2}]

Why not the other options?

- Option 1 (\frac{\Delta u}{u}+\frac{\Delta v}{v}) would be the relative error formula for a product, not applicable here.
- Options 3 and 4 introduce incorrect prefactors (2f or f) that do not follow from the correct differentiation of the lens formula.
- Only Option 2 correctly captures the f² factor that emerges from converting the error in \frac{1}{f} to the error in f, combined with the u² and v² denominators from differentiating the reciprocals.

This result is physically meaningful: a larger focal length f leads to a much larger absolute error in its measurement (it grows as f²), which explains why measuring focal lengths of very powerful (long focal length) lenses is inherently less precise with the same measuring instruments.