A lens forms a virtual, diminished image of an object placed at 2m from it. The size of image is half of the object. Which one of the following statements is correct regarding the nature and focal length of the lens?

Options

A. Concave, `|f|=1m`
B. Convex, `|f|=1m`
C. Concave, `|f|=2m`
D. Convex, `|f|=2m`

Explanation

To solve the problem step by step, we will analyze the information given and apply the lens formula and magnification concepts.

### Step 1: Understand the Given Information
We have a lens that forms a virtual, diminished image of an object placed at a distance of 2 m from the lens. The size of the image is half that of the object.

### Step 2: Assign Values
- The object distance (u) is given as -2 m (the negative sign indicates that the object is placed on the same side as the incoming light).
- The magnification (m) is given as the ratio of the height of the image (h') to the height of the object (h). Since the image is half the size of the object, we have:
\[ m = \frac{h'}{h} = \frac{1}{2} \]

### Step 3: Relate Magnification to Image Distance
The magnification can also be expressed in terms of image distance (v) and object distance (u):
\[ m = -\frac{v}{u} \]
Substituting the values we have:
\[ \frac{1}{2} = -\frac{v}{-2} \]
This simplifies to:
\[ \frac{1}{2} = \frac{v}{2} \]
Cross-multiplying gives:
\[ v = 1 \text{ m} \]

### Step 4: Use the Lens Formula
The lens formula relates the object distance (u), image distance (v), and focal length (f):
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of v and u:
\[ \frac{1}{f} = \frac{1}{1} - \frac{1}{-2} \]
This simplifies to:
\[ \frac{1}{f} = 1 + \frac{1}{2} = \frac{3}{2} \]
Thus,
\[ f = \frac{2}{3} \text{ m} \approx 0.67 \text{ m} \]

### Step 5: Determine the Nature of the Lens
Since the focal length (f) is positive, this indicates that the lens is a **convex lens**.

### Conclusion
The correct statement regarding the nature and focal length of the lens is that it is a **convex lens** with a focal length of approximately **0.67 m**.

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