An object is placed at a distance of `f//2` from a convex lens. The image will be

Options

A. At one of the foci, virtual and double its size
B. At 3f/2, real and inverted
C. At 2f, virtual and erect
D. At f, real and inverted

Answer

To solve the problem of finding the characteristics of the image formed by a convex lens when an object is placed at a distance of \( \frac{f}{2} \) from the lens, we can follow these steps:

### Step 1: Understand the Lens Formula
The lens formula for a convex lens is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where:
- \( f \) = focal length of the lens (positive for a convex lens)
- \( v \) = image distance (positive if the image is on the opposite side of the object)
- \( u \) = object distance (negative as per the sign convention)

### Step 2: Assign Values
Given that the object is placed at a distance of \( \frac{f}{2} \):
- \( u = -\frac{f}{2} \) (negative because the object is on the same side as the incoming light)

### Step 3: Substitute Values into the Lens Formula
Substituting \( u \) into the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{-\frac{f}{2}} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{v} + \frac{2}{f} \]

### Step 4: Rearranging the Equation
Rearranging the equation to isolate \( \frac{1}{v} \):
\[ \frac{1}{v} = \frac{1}{f} - \frac{2}{f} = -\frac{1}{f} \]

### Step 5: Solve for \( v \)
Taking the reciprocal gives:
\[ v = -f \]
This indicates that the image is formed at a distance of \( f \) on the same side as the object.

### Step 6: Determine the Nature of the Image
Since \( v \) is negative, the image is virtual and formed on the same side as the object.

### Step 7: Calculate the Magnification
The magnification \( m \) is given by:
\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]
Substituting the values:
\[ m = \frac{-f}{-\frac{f}{2}} = 2 \]
This means the image is twice the height of the object.

### Step 8: Conclusion
The image is virtual, erect, and double the size of the object, formed at a distance \( f \) from the lens.

### Final Answer
The image will be virtual, erect, and at a distance of \( f \) from the lens, with a height that is double the height of the object.

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