At what distance from a convex lens of focal length `30cm` an object should be placed so that the size of image be `(1)/(4) `that of object?

Options

A. 30 cm
B. 60 cm
C. 15 cm
D. 150 cm

Explanation

To solve the problem, we need to determine the distance at which an object should be placed from a convex lens so that the size of the image is one-fourth that of the object.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Focal length of the convex lens, \( f = +30 \, \text{cm} \) (positive because it's a convex lens).
- Size of the image is one-fourth that of the object, which implies the magnification \( m = \frac{h'}{h} = \frac{1}{4} \).

2. **Determine the Type of Image**:
- Since the image size is smaller than the object, the image is real and inverted. Therefore, the magnification is given by:
\[ m = -\frac{v}{u} \]
- Here, \( v \) is the image distance and \( u \) is the object distance.

3. **Express Image Distance in Terms of Object Distance**:
- From the magnification formula, we have:
\[ -\frac{v}{u} = \frac{1}{4} \]
- Rearranging gives:
\[ v = -\frac{1}{4} u \]

4. **Use the Lens Formula**:
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Substituting \( f = 30 \, \text{cm} \) and \( v = -\frac{1}{4} u \):
\[ \frac{1}{30} = \frac{1}{-\frac{1}{4}u} - \frac{1}{u} \]

5. **Simplify the Equation**:
- The term \( \frac{1}{-\frac{1}{4}u} \) simplifies to \( -\frac{4}{u} \):
\[ \frac{1}{30} = -\frac{4}{u} - \frac{1}{u} \]
- Combine the terms on the right:
\[ \frac{1}{30} = -\frac{5}{u} \]

6. **Solve for \( u \)**:
- Rearranging gives:
\[ \frac{5}{u} = -\frac{1}{30} \]
- Cross-multiplying results in:
\[ 5 \cdot 30 = -u \]
- Thus:
\[ u = -150 \, \text{cm} \]

7. **Determine the Object Distance**:
- Since \( u \) is negative, it indicates that the object is placed 150 cm in front of the lens.

### Final Answer:
The object should be placed at a distance of **150 cm** from the convex lens.

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