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. `(p)/(2)((1)/(u)+(1)/(v))`
B. `p((1)/(u)+(1)/(v))`
C. `(p)/(2)((1)/(u)-(1)/(v))`
D. `p((1)/(u)-(1)/(v))`

Answer

To solve the problem, we need to find the relative error in the focal length \( f \) given the relationship between \( u \), \( v \), and \( f \) as:

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

### Step 1: Differentiate the equation
We start by differentiating both sides of the equation with respect to the variables \( u \) and \( v \).

\[ \frac{d}{du}\left(\frac{1}{v}\right) - \frac{d}{du}\left(\frac{1}{u}\right) = \frac{d}{du}\left(\frac{2}{f}\right) \]

Using the chain rule, we have:

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

This simplifies to:

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

### Step 2: Substitute the errors
Given that the errors in measuring \( u \) and \( v \) are equal and denoted by \( \alpha \), we have:

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

Substituting these into the differentiated equation gives:

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

### Step 3: Factor out \( \alpha \)
Factoring out \( \alpha \) from the left side:

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

### Step 4: Rearranging for \( df \)
Rearranging gives:

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

### Step 5: Calculate the relative error in \( f \)
The relative error in \( f \) is given by:

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

This simplifies to:

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

### Final Expression
Thus, the relative error in \( f \) is:

\[ \text{Relative Error in } f = \frac{\alpha}{2} \left(\frac{1}{v^2} + \frac{1}{u^2}\right) \]