Step-by-Step Solution

1. Determine the Least Count (LC) of the Scale

The optical bench scale has four equal divisions in each cm. This means the value of one division is: $1 \text{ division} = \frac{1 \text{ cm}}{4} = 0.25 \text{ cm}$ The least count (LC) of the instrument is the smallest measurement it can make, which is 0.25 cm. The error in any single position measurement is taken as the least count. $\Delta x = \text{LC} = 0.25 \text{ cm}$

2. Calculate Object Distance (u) and Image Distance (v)

The positions of the components on the optical bench are given:

• Position of object pin, $x_O = 45$ cm
• Position of lens, $x_L = 75$ cm
• Position of image pin, $x_I = 135$ cm

The object distance ($u$) is the distance between the object pin and the lens. $u = |x_L - x_O| = |75 - 45| = 30 \text{ cm}$ The image distance ($v$) is the distance between the image pin and the lens. $v = |x_I - x_L| = |135 - 75| = 60 \text{ cm}$

3. Calculate the Focal Length (f) of the Lens

Using the lens formula, and noting that for a convex lens forming a real image, both $u$ and $v$ are positive when we consider their magnitudes: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ Substituting the values of $u$ and $v$: $\frac{1}{f} = \frac{1}{60} + \frac{1}{30} = \frac{1 + 2}{60} = \frac{3}{60} = \frac{1}{20}$ $f = 20 \text{ cm}$

4. Determine the Error in u and v

The object distance $u$ is calculated from two position measurements, $x_L$ and $x_O$. The maximum possible error in $u$ is the sum of the errors in these individual measurements. $\Delta u = \Delta x_L + \Delta x_O = 0.25 \text{ cm} + 0.25 \text{ cm} = 0.5 \text{ cm}$ Similarly, the image distance $v$ is calculated from $x_I$ and $x_L$. The maximum possible error in $v$ is: $\Delta v = \Delta x_I + \Delta x_L = 0.25 \text{ cm} + 0.25 \text{ cm} = 0.5 \text{ cm}$

5. Calculate the Percentage Error in the Focal Length

To find the error in focal length, we start with the lens formula and use error propagation. Differentiating the lens formula: $-\frac{df}{f^2} = -\frac{du}{u^2} - \frac{dv}{v^2}$ For maximum error, we consider the absolute values: $\frac{\Delta f}{f^2} = \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}$ The relative error in $f$ is: $\frac{\Delta f}{f} = f \left( \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2} \right)$ Substituting the calculated values: $\frac{\Delta f}{f} = 20 \left( \frac{0.5}{30^2} + \frac{0.5}{60^2} \right)$ $\frac{\Delta f}{f} = 20 \times 0.5 \left( \frac{1}{900} + \frac{1}{3600} \right)$ $\frac{\Delta f}{f} = 10 \left( \frac{4+1}{3600} \right) = 10 \left( \frac{5}{3600} \right) = \frac{50}{3600} = \frac{1}{72}$ The percentage error is the relative error multiplied by 100: $\text{Percentage Error} = \frac{\Delta f}{f} \times 100 = \frac{1}{72} \times 100 = \frac{100}{72} = \frac{25}{18}$ Converting the fraction to a decimal: $\text{Percentage Error} = 1.3888... \%$

6. Final Answer and Comparison

The calculated percentage error is $1.3888...\%$. Rounding this to two decimal places gives $1.39\%$. The stored answer is $1.38$. The discrepancy is likely due to the stored answer being truncated to two decimal places instead of being rounded. Following standard rounding conventions, the answer should be $1.39$.