A convex lens has a focal length of 20 cm. It is used to form an image of an object placed 15 cm from lens. The image is

Options

A. virtual inverted and enlarge
B. real,inverted and diminished
C. real,inverted and enlarge
D. virtual,erect and enlarge

Explanation

To solve the problem step-by-step, we will use the lens formula and the properties of the image formed by a convex lens.

### Step 1: Identify the given values
- Focal length (f) of the convex lens = +20 cm (positive because it is a convex lens)
- Object distance (u) = -15 cm (negative because object distance is measured against the direction of incident light)

### Step 2: Use 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
- \( v \) = image distance
- \( u \) = object distance

### Step 3: Substitute the known values into the lens formula
Substituting the values we have into the lens formula:
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{-15} \]
This simplifies to:
\[ \frac{1}{20} = \frac{1}{v} + \frac{1}{15} \]

### Step 4: Find a common denominator and solve for \( v \)
To solve for \( v \), we first find a common denominator for the fractions on the right side. The common denominator of 20 and 15 is 60.
\[ \frac{1}{20} = \frac{3}{60} \quad \text{and} \quad \frac{1}{15} = \frac{4}{60} \]
Substituting these into the equation gives:
\[ \frac{3}{60} = \frac{1}{v} + \frac{4}{60} \]
Now, rearranging gives:
\[ \frac{1}{v} = \frac{3}{60} - \frac{4}{60} = -\frac{1}{60} \]
Thus,
\[ v = -60 \, \text{cm} \]

### Step 5: Determine the properties of the image
1. **Nature of the image**: Since \( v \) is negative, the image is virtual.
2. **Orientation of the image**: The image formed by a convex lens when the object is between the focal point and the lens is erect.
3. **Size of the image**: The magnification (m) can be calculated using the formula:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{-60}{-15} = 4 \]
This indicates that the image is enlarged (4 times the size of the object).

### Final Conclusion
The image formed is virtual, erect, and enlarged.

### Summary of the properties of the image:
- **Virtual**
- **Erect**
- **Enlarged**