In a given optical bench, a needle of length 10 cm is used to estimate bench error. The object needle, image neddle & lens holder have their reading as shown. `X_(0) = 1.1 cm , X_(1) = 0.8 cm , X_(L) = 10.9 cm`
Estimate the bench error which are present in image neddle holder and object needle holder. Also find the focal length of the convex lens when `X_(0)= 0.6 cm , X_(1) = 22.5 cm , X_(L) = 11.4 cm`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the Bench Error
The bench error can be calculated using the formula:
\[ \text{Bench Error} = (X_1 - X_0) - \text{Length of the Needle} \]

Given:
- \( X_0 = 1.1 \, \text{cm} \)
- \( X_1 = 0.8 \, \text{cm} \)
- Length of the Needle = 10 cm

Substituting the values:
\[ \text{Bench Error} = (0.8 \, \text{cm} - 1.1 \, \text{cm}) - 10 \, \text{cm} \]
\[ \text{Bench Error} = (-0.3 \, \text{cm}) - 10 \, \text{cm} \]
\[ \text{Bench Error} = -10.3 \, \text{cm} \]

### Step 2: Calculate the Focal Length of the Convex Lens
To find the focal length \( F \) of the convex lens, we will use the lens formula:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]

Where:
- \( U = X_L - X_0 \)
- \( V = X_1 - X_0 \)

Given:
- \( X_0 = 0.6 \, \text{cm} \)
- \( X_1 = 22.5 \, \text{cm} \)
- \( X_L = 11.4 \, \text{cm} \)

Calculating \( U \) and \( V \):
1. Calculate \( U \):
\[ U = X_L - X_0 = 11.4 \, \text{cm} - 0.6 \, \text{cm} = 10.8 \, \text{cm} \]

2. Calculate \( V \):
\[ V = X_1 - X_0 = 22.5 \, \text{cm} - 0.6 \, \text{cm} = 21.9 \, \text{cm} \]

Now substituting \( U \) and \( V \) into the lens formula:
\[ \frac{1}{F} = \frac{1}{21.9} - \frac{1}{10.8} \]

Calculating the right-hand side:
\[ \frac{1}{F} = \frac{10.8 - 21.9}{21.9 \times 10.8} \]
\[ \frac{1}{F} = \frac{-11.1}{236.52} \]
\[ F \approx -21.3 \, \text{cm} \]

### Step 3: Calculate the Error in Focal Length
To find the error in focal length (\( \Delta F \)), we can use the formula:
\[ \Delta F = \sqrt{\left(\frac{\Delta U}{U^2}\right) + \left(\frac{\Delta V}{V^2}\right)} \]

Assuming \( \Delta U = 0.1 \, \text{cm} \) and \( \Delta V = 0.1 \, \text{cm} \):
\[ \Delta F = \sqrt{\left(\frac{0.1}{10.8^2}\right) + \left(\frac{0.1}{21.9^2}\right)} \]
Calculating:
\[ \Delta F = \sqrt{\left(\frac{0.1}{116.64}\right) + \left(\frac{0.1}{480.81}\right)} \]
\[ \Delta F = \sqrt{0.000856 + 0.000208} \]
\[ \Delta F \approx \sqrt{0.001064} \]
\[ \Delta F \approx 0.0327 \, \text{cm} \]

### Final Result
The bench error is \( -10.3 \, \text{cm} \) and the focal length of the lens is approximately \( -21.3 \, \text{cm} \) with an error of \( \Delta F \approx 0.0327 \, \text{cm} \).

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