A convex mirror of focal length f produced an image `(1//n)^(th)` of the size of the object. The distance of the object from the mirror is

Options

A. nf
B. `f//n`
C. `(n+1)f`
D. `(n-1)f`

Explanation

To solve the problem step by step, we will use the mirror formula and the magnification concept for a convex mirror.

### Step 1: Understand the given information
We have a convex mirror with a focal length \( f \) and an image that is \( \frac{1}{n} \) times the size of the object.

### Step 2: Define the variables
Let:
- \( u \) = distance of the object from the mirror (which will be negative for real objects in the mirror convention).
- \( v \) = distance of the image from the mirror (which will be positive for virtual images in the mirror convention).
- The magnification \( m \) is given by \( m = \frac{h'}{h} = \frac{v}{u} \), where \( h' \) is the height of the image and \( h \) is the height of the object.

### Step 3: Set up the magnification equation
Since the image is \( \frac{1}{n} \) times the size of the object, we can write:
\[ m = \frac{1}{n} = \frac{v}{u} \]
From this, we can express \( v \) in terms of \( u \):
\[ v = \frac{u}{n} \]

### Step 4: Substitute the values into the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting \( v = \frac{u}{n} \) into the mirror formula:
\[ \frac{1}{f} = \frac{1}{\frac{u}{n}} + \frac{1}{u} \]
This simplifies to:
\[ \frac{1}{f} = \frac{n}{u} + \frac{1}{u} \]
Combining the fractions on the right side:
\[ \frac{1}{f} = \frac{n + 1}{u} \]

### Step 5: Rearranging to find \( u \)
Now, rearranging the equation to solve for \( u \):
\[ u = (n + 1)f \]

### Step 6: Apply the sign convention
Since we defined \( u \) as negative (as per the sign convention for mirrors), we have:
\[ u = -(n + 1)f \]

### Final Answer
The distance of the object from the mirror is:
\[ u = - (n + 1)f \]