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 1: Understand the assertion
The assertion states that when a convex lens and a concave lens are kept in contact, they will behave as a diverging lens if the focal length of the convex lens is more.

### Step 2: Use the lens formula
The lens formula for two lenses in contact is given by:
\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]
where \( f \) is the equivalent focal length, \( f_1 \) is the focal length of the convex lens (positive), and \( f_2 \) is the focal length of the concave lens (negative).

### Step 3: Assign values
Let’s assume:
- \( f_1 = f \) (focal length of the convex lens, positive)
- \( f_2 = -f' \) (focal length of the concave lens, negative)

### Step 4: Analyze the condition
If the focal length of the convex lens \( f_1 \) is more than the absolute value of the focal length of the concave lens \( |f_2| \), we can express this as:
\[ f_1 > |f_2| \]
This means:
\[ f > f' \]

### Step 5: Substitute in the lens formula
Substituting the values in the lens formula:
\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{f} + \frac{1}{-f'} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{f} - \frac{1}{f'} \]

### Step 6: Determine the sign of \( f \)
If \( f_1 > |f_2| \), the term \( \frac{1}{f_2} \) (which is negative) will dominate, leading to:
\[ \frac{1}{f} < 0 \implies f < 0 \]
This indicates that the equivalent focal length \( f \) is negative, meaning the combination behaves as a diverging lens.

### Step 7: Analyze the reason
The reason states that the power of a concave lens is always less than that of a convex lens. The power \( P \) of a lens is given by:
\[ P = \frac{1}{f} \]
For a convex lens, \( P_1 > 0 \) (positive), and for a concave lens, \( P_2 < 0 \) (negative). Thus, the power of the concave lens is indeed less than that of the convex lens in terms of numerical value.

### Step 8: Conclusion
- The assertion is **true** because the combination behaves as a diverging lens when the focal length of the convex lens is greater.
- The reason is **false** because it incorrectly implies a comparison of powers without acknowledging that the sign convention does not affect the absolute values.

### Final Answer:
- **Assertion**: True
- **Reason**: False