1. Given data

• Least count of scale $=0.2\,\text{cm}$
• Object position $=80\,\text{cm}$
• Lens position $=100\,\text{cm}$
• Image pin position $=180\,\text{cm}$

So, $u = 100-80 = 20\,\text{cm}$ $v = 180-100 = 80\,\text{cm}$

2. Focal length of convex lens

Using lens formula in magnitude form, $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ $f=\frac{uv}{u+v}$

Substitute values: $f=\frac{20\times 80}{20+80}=\frac{1600}{100}=16\,\text{cm}$

3. Error in measured distances

Each position measured on the scale has maximum error equal to least count: $\Delta x = 0.2\,\text{cm}$

Now, $u = x_L-x_O$ So maximum error in $u$ is $\Delta u = \Delta x_L + \Delta x_O = 0.2+0.2=0.4\,\text{cm}$

Similarly, $v = x_I-x_L$ So maximum error in $v$ is $\Delta v = \Delta x_I + \Delta x_L = 0.2+0.2=0.4\,\text{cm}$

Thus, $\frac{\Delta u}{u}=\frac{0.4}{20}=0.02$ $\frac{\Delta v}{v}=\frac{0.4}{80}=0.005$

4. Propagation of error in focal length

Since $f=\frac{uv}{u+v}$ Taking logarithmic differentiation, $\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}+\frac{\Delta (u+v)}{u+v}$ But because $f=uv/(u+v)$, the denominator contributes with minus sign: $\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}+\frac{\Delta(u+v)}{u+v} \text{ is incorrect}$ Correctly, $\ln f = \ln u + \ln v - \ln (u+v)$ Hence, $\frac{\Delta f}{f}=\frac{\Delta u}{u}+\frac{\Delta v}{v}+\frac{\Delta(u+v)}{u+v} \text{ with sign taken in magnitude form}$ For maximum error, $\Delta(u+v)=\Delta u+\Delta v=0.4+0.4=0.8\,\text{cm}$

Therefore, $\frac{\Delta f}{f}=\frac{0.4}{20}+\frac{0.4}{80}+\frac{0.8}{100}$ $=0.02+0.005+0.008=0.033$

This gives $3.3\%$, which is not among the options, so let us use the standard differential form directly.

From $f=\frac{uv}{u+v}$ Differentiate: $df=\frac{v^2\,du+u^2\,dv}{(u+v)^2}$

So maximum fractional error is $\frac{\Delta f}{f}=\frac{v^2\Delta u+u^2\Delta v}{uv(u+v)}$

Substitute $u=20$, $v=80$, $\Delta u=\Delta v=0.4$: $\frac{\Delta f}{f}=\frac{80^2(0.4)+20^2(0.4)}{20\cdot 80\cdot 100}$ $=\frac{0.4(6400+400)}{160000}$ $=\frac{2720}{160000}=0.017$

Thus, $\%\text{ error}=0.017\times 100=1.70\%$

5. Final answer

The percentage error in focal length is $\boxed{1.70\%}$ So, Option A is correct.