A thin convex lens of focal length 10 cm and a thin concave lens of focal length 26.2 cm are in contact. The combination acts as

Options

A. concave lens of focal length 16.4 cm
B. convex lens of focal length 16.2 cm
C. concave or convex lens depends upon `mu`of material of lenses
D. none of the above.

Answer

To determine the nature of the combination of a thin convex lens and a thin concave lens that are in contact, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Focal Lengths:**
- The focal length of the convex lens (F1) is given as +10 cm.
- The focal length of the concave lens (F2) is given as -26.2 cm (negative because it is a concave lens).

2. **Convert Focal Lengths to Meters (if necessary):**
- F1 = 10 cm = 0.1 m
- F2 = -26.2 cm = -0.262 m

3. **Calculate the Power of Each Lens:**
- The power (P) of a lens is given by the formula:
\[ P = \frac{1}{F} \]
- For the convex lens:
\[ P_1 = \frac{1}{F1} = \frac{1}{0.1} = 10 \, \text{D} \]
- For the concave lens:
\[ P_2 = \frac{1}{F2} = \frac{1}{-0.262} \approx -3.81 \, \text{D} \]

4. **Calculate the Total Power of the Combination:**
- The total power (P) of the combination is the sum of the powers of the individual lenses:
\[ P = P_1 + P_2 \]
\[ P = 10 - 3.81 = 6.19 \, \text{D} \]

5. **Determine the Focal Length of the Combination:**
- The focal length (F) of the combination can be found using the power:
\[ P = \frac{1}{F} \]
- Rearranging gives:
\[ F = \frac{1}{P} = \frac{1}{6.19} \approx 0.161 \, \text{m} \]
- Converting to centimeters:
\[ F \approx 16.1 \, \text{cm} \]

6. **Determine the Nature of the Combination:**
- Since the focal length is positive, the combination acts as a **convex lens**.

### Final Answer:
The combination of a thin convex lens of focal length 10 cm and a thin concave lens of focal length 26.2 cm acts as a **convex lens** with a focal length of approximately 16.1 cm.