A convex lens of focal length `f` produces a virtual image `n` times the size of the object. Then the distance of the object from the lens is

Options

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

Explanation

To solve the problem of finding the distance of the object from a convex lens that produces a virtual image `n` times the size of the object, we can follow these steps:

### Step 1: Understand the Magnification
The magnification (m) produced by a lens is given by the formula:
\[ m = \frac{h'}{h} = \frac{v}{u} \]
where:
- \( h' \) is the height of the image,
- \( h \) is the height of the object,
- \( v \) is the image distance,
- \( u \) is the object distance.

Since the image is virtual and erect, the magnification is positive. Thus, we have:
\[ m = n \]

### Step 2: Relate Magnification to Object and Image Distances
From the magnification formula, we can express it as:
\[ n = \frac{v}{u} \]
This implies:
\[ v = n \cdot u \]

### Step 3: Use the Lens Formula
The lens formula for a convex lens is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging this gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]

### Step 4: Substitute for v
Substituting \( v = n \cdot u \) into the lens formula:
\[ \frac{1}{n \cdot u} = \frac{1}{f} + \frac{1}{u} \]

### Step 5: Clear the Denominators
To eliminate the fractions, multiply through by \( n \cdot u \cdot f \):
\[ f = n + n \cdot u \]

### Step 6: Rearranging for u
Rearranging gives:
\[ n \cdot u = f - n \]
Thus:
\[ u = \frac{f - n}{n} \]

### Step 7: Consider the Sign of u
Since the object is on the same side as the incoming light for a convex lens, the object distance \( u \) is taken as negative in the lens convention. Therefore:
\[ u = -\frac{f - n}{n} \]

### Step 8: Final Expression for Object Distance
Thus, the distance of the object from the lens is:
\[ |u| = \frac{n - 1}{n} \cdot f \]

### Conclusion
The distance of the object from the lens is:
\[ \text{Distance} = \frac{n - 1}{n} \cdot f \]