An object has image thrice of its original size when kept at `8 cm` and `16 cm` from a convex lens. Focal length of the lens is

Options

A. less than 8 cm
B. 8 cm
C. 16 cm
D. between 8 and 16 cm

Explanation

To find the focal length of the convex lens given that the image is thrice the size of the object when placed at distances of 8 cm and 16 cm, we can use the lens formula and the magnification formula.

### Step-by-Step Solution:

1. **Understand the Magnification Formula:**
The magnification (M) produced by a lens is given by:
\[ M = \frac{V}{U} \]
where \( V \) is the image distance and \( U \) is the object distance. Given that the image is thrice the size of the object, we have:
\[ M = 3 \]

2. **Set Up the Object Distances:**
For the two cases, the object distances are:
- Case 1: \( U_1 = -8 \, \text{cm} \) (negative because the object is on the same side as the incoming light)
- Case 2: \( U_2 = -16 \, \text{cm} \)

3. **Apply the Magnification Formula for Both Cases:**
For Case 1:
\[ 3 = \frac{V_1}{-8} \]
Rearranging gives:
\[ V_1 = -24 \, \text{cm} \]

For Case 2:
\[ 3 = \frac{V_2}{-16} \]
Rearranging gives:
\[ V_2 = -48 \, \text{cm} \]

4. **Use the Lens Formula:**
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
We will apply this for both cases.

**For Case 1:**
\[ \frac{1}{F} = \frac{1}{-24} - \frac{1}{-8} \]
Simplifying:
\[ \frac{1}{F} = -\frac{1}{24} + \frac{3}{24} = \frac{2}{24} = \frac{1}{12} \]
Therefore, \( F = 12 \, \text{cm} \).

**For Case 2:**
\[ \frac{1}{F} = \frac{1}{-48} - \frac{1}{-16} \]
Simplifying:
\[ \frac{1}{F} = -\frac{1}{48} + \frac{3}{48} = \frac{2}{48} = \frac{1}{24} \]
Therefore, \( F = 12 \, \text{cm} \).

5. **Conclusion:**
The focal length of the convex lens is:
\[ F = 12 \, \text{cm} \]