Assertion:A convex lens and a concave lens are kept in contact. They will behave as a diverging lens if focal length of convex lens is more.
Reason: Power of a concave lens is always less than the power of a convex lens, as power of concave lens is negative whereas power of convex lens is positive.

Options

A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
B. If both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.
C. If both Assertion is true, but the Reason is false.
D. If both Assertion is false but the Reason is true.

Explanation

To solve the question, we need to analyze both the assertion and the reason provided.

### Step-by-Step Solution:

1. **Understanding the Assertion**:
- The assertion states that a convex lens and a concave lens kept in contact will behave as a diverging lens if the focal length of the convex lens is more.
- A convex lens has a positive focal length (f1) and a concave lens has a negative focal length (f2).

2. **Understanding the Focal Lengths**:
- Let’s denote the focal length of the convex lens as \( f_1 \) (positive) and the focal length of the concave lens as \( f_2 \) (negative).
- For the system of lenses to behave as a diverging lens, the net focal length (F) must be negative.

3. **Using the Lens Formula**:
- The lens formula for two lenses in contact is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
- Since \( f_2 \) is negative, we can rewrite the equation as:
\[ \frac{1}{F} = \frac{1}{f_1} - \frac{1}{|f_2|} \]

4. **Condition for Diverging Lens**:
- For the combined lens to behave as a diverging lens, we need:
\[ \frac{1}{F} < 0 \]
- This implies:
\[ \frac{1}{f_1} < \frac{1}{|f_2|} \]
- Rearranging gives:
\[ f_1 > |f_2| \]
- This means that the focal length of the convex lens must be greater than the magnitude of the focal length of the concave lens.

5. **Conclusion on the Assertion**:
- Since the assertion states that the focal length of the convex lens is more, it is indeed true that \( f_1 > |f_2| \) leads to the system behaving as a diverging lens.

6. **Analyzing the Reason**:
- The reason states that the power of a concave lens is always less than the power of a convex lens.
- Power (P) is defined as:
\[ P = \frac{1}{f} \]
- For a convex lens, the power is positive, and for a concave lens, the power is negative.
- While it is true that the power of a concave lens is negative and the power of a convex lens is positive, saying that one is "less" than the other is misleading because they are of different signs.

7. **Conclusion on the Reason**:
- Therefore, the reason is incorrect because it misrepresents the relationship between the powers of the lenses.

### Final Answer:
- The assertion is true, but the reason is false.