An object is placed 12 cm from a convex lens whose focal length is 10cm. The image must be:

Options

A. virtual and enlarged
B. virtual and reduced in size
C. real and reduced in size
D. real and enlarged

Explanation

To determine the characteristics of the image formed by a convex lens when an object is placed at a certain distance, we can use the lens formula and magnification concepts. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Object distance (u) = -12 cm (the object distance is taken as negative in lens formula)
- Focal length (f) = +10 cm (the focal length of a convex lens is positive)

### Step 2: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) = focal length of the lens
- \( v \) = image distance from the lens
- \( u \) = object distance from the lens

### Step 3: Substitute the known values into the lens formula
Substituting the values we have:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-12} \]
This simplifies to:
\[ \frac{1}{10} = \frac{1}{v} + \frac{1}{12} \]

### Step 4: Find a common denominator and solve for \( v \)
To solve for \( v \), we first find a common denominator for the fractions:
The common denominator of 10 and 12 is 60.
Rewriting the equation:
\[ \frac{6}{60} = \frac{1}{v} + \frac{5}{60} \]
Now, subtract \(\frac{5}{60}\) from both sides:
\[ \frac{6}{60} - \frac{5}{60} = \frac{1}{v} \]
\[ \frac{1}{60} = \frac{1}{v} \]
Thus, \( v = 60 \) cm.

### Step 5: Determine the nature of the image
Since \( v \) is positive, the image is real and located on the opposite side of the lens from the object.

### Step 6: Calculate the magnification
The magnification (m) is given by:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
Where:
- \( h' \) = height of the image
- \( h \) = height of the object

Substituting the values:
\[ m = -\frac{60}{-12} = 5 \]
This indicates that the image is enlarged (5 times the height of the object).

### Final Conclusion
The image formed by the convex lens is **real and enlarged**.

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