A lens form a diminshed and erect image of an object. The magnifiation is `(1)/(4)`. Find ratio of distances between object and focus and focus and image:

Options

A. `4:1`
B. `1:4`
C. `8:1`
D. `2:1`

Explanation

To solve the problem, we need to find the ratio of distances between the object and the focus, and the focus and the image, given that the lens forms a diminished and erect image with a magnification of \( \frac{1}{4} \).

### Step 1: Understand the Magnification
The magnification \( m \) is given by the formula:
\[ m = \frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance. Since the image is diminished and erect, the magnification is positive and less than 1, which is given as \( m = \frac{1}{4} \).

### Step 2: Relate Image and Object Distances
From the magnification formula, we can express the relationship between the image distance \( v \) and the object distance \( u \):
\[ \frac{v}{u} = \frac{1}{4} \]
Assuming the image distance \( v = x \), we can express the object distance \( u \) as:
\[ u = 4v = 4x \]

### Step 3: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting \( v = x \) and \( u = -4x \) (negative because the object is on the same side as the incoming light for a concave lens):
\[ \frac{1}{f} = \frac{1}{x} - \frac{1}{-4x} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{x} + \frac{1}{4x} = \frac{4 + 1}{4x} = \frac{5}{4x} \]
Thus, we find:
\[ f = \frac{4x}{5} \]

### Step 4: Calculate Distances
Now we need to find the distances between the object and the focus, and the focus and the image.

1. **Distance between Object and Focus**:
- The object is at \( -4x \) and the focus is at \( -\frac{4x}{5} \).
- The distance \( d_{object-focus} \) is:
\[ d_{object-focus} = \left| -4x - \left(-\frac{4x}{5}\right) \right| = \left| -4x + \frac{4x}{5} \right| = \left| -\frac{20x}{5} + \frac{4x}{5} \right| = \left| -\frac{16x}{5} \right| = \frac{16x}{5} \]

2. **Distance between Focus and Image**:
- The focus is at \( -\frac{4x}{5} \) and the image is at \( -x \).
- The distance \( d_{focus-image} \) is:
\[ d_{focus-image} = \left| -\frac{4x}{5} - (-x) \right| = \left| -\frac{4x}{5} + x \right| = \left| -\frac{4x}{5} + \frac{5x}{5} \right| = \left| \frac{x}{5} \right| = \frac{x}{5} \]

### Step 5: Find the Ratio
Now we can find the ratio of the distances:
\[ \text{Ratio} = \frac{d_{object-focus}}{d_{focus-image}} = \frac{\frac{16x}{5}}{\frac{x}{5}} = \frac{16x}{5} \cdot \frac{5}{x} = 16 \]
Thus, the ratio of the distances between the object and focus, and the focus and image is:
\[ \text{Ratio} = 16 : 1 \]

### Final Answer
The ratio of distances between the object and focus, and the focus and image is \( 16 : 1 \).