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.

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Assertion is false and Reason is true.

Assertion is True and Reason is false.

Both Assertion and Reason are correct and R is not correct explanation of A

Both Assertion and Reason are correct and R is correct explanation of A

answer is C.

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Detailed Solution

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$f_{1} =$ Focal length of convex lens (+ve)
$f_{2} =$ Focal length of concave lens (-ve)
If $f_{1} > | f_{2} |,$ then F comes out to be negative So, it becomes a diverging lens
Negative power of a diverging lens is just for its diverging lens, nothing else