The focal length of a mirror is given by `1/v-1/u=2/f`. If equal errors `alpha` are made inmeasuring `u and v`. Then relative error in `f` is

Options

A. `2/a`
B. ``a(1/u+1/v)`
C. `a(1/u-1/v)`
D. none of these

Explanation

To solve the problem, we start with the equation given for the focal length of a mirror:

\[ \frac{1}{v} - \frac{1}{u} = \frac{2}{f} \]

### Step 1: Differentiate the equation
We will differentiate both sides of the equation with respect to \( u \) and \( v \). The left-hand side involves the differentiation of fractions, and we apply the product rule.

Differentiating the left-hand side:

\[ d\left(\frac{1}{v}\right) - d\left(\frac{1}{u}\right) = 0 \]

Using the chain rule, we have:

\[ -\frac{1}{v^2} dv + \frac{1}{u^2} du = 0 \]

Differentiating the right-hand side:

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

### Step 2: Set the derivatives equal
Now we set the differentiated left-hand side equal to the differentiated right-hand side:

\[ -\frac{1}{v^2} dv + \frac{1}{u^2} du = -\frac{2}{f^2} df \]

### Step 3: Substitute equal errors
According to the problem, equal errors \( \alpha \) are made in measuring \( u \) and \( v \). Therefore, we can set:

\[ du = \alpha \quad \text{and} \quad dv = \alpha \]

Substituting these into the equation gives us:

\[ -\frac{1}{v^2} \alpha + \frac{1}{u^2} \alpha = -\frac{2}{f^2} df \]

### Step 4: Factor out \( \alpha \)
Factoring out \( \alpha \) from the left-hand side:

\[ \alpha \left(-\frac{1}{v^2} + \frac{1}{u^2}\right) = -\frac{2}{f^2} df \]

### Step 5: Rearranging for \( df \)
Rearranging gives:

\[ df = -\frac{f^2}{2} \left(-\frac{1}{v^2} + \frac{1}{u^2}\right) \alpha \]

### Step 6: Find the relative error in \( f \)
The relative error in \( f \) is given by:

\[ \frac{df}{f} \]

Substituting \( df \) into this expression:

\[ \frac{df}{f} = \frac{-\frac{f^2}{2} \left(-\frac{1}{v^2} + \frac{1}{u^2}\right) \alpha}{f} \]

This simplifies to:

\[ \frac{df}{f} = \frac{f}{2} \left(\frac{1}{v^2} - \frac{1}{u^2}\right) \alpha \]

### Step 7: Final expression for relative error
Thus, the relative error in \( f \) is:

\[ \text{Relative error in } f = \alpha \left(\frac{1}{v} + \frac{1}{u}\right) \]

### Final Answer
The relative error in \( f \) is:

\[ \text{Relative error in } f = \alpha \left(\frac{1}{v} + \frac{1}{u}\right) \]