Question

Convex lens of focal length \(f_1\) is placed in contact with a concave lens of focal length \(f_2\). For \(f_1 > f_2\) combination will behave:

• A. like a concave lens
• B. like a convex lens
• C. like a plane glass slab
• D. none of these

Hint:

Use \(\frac{1}{F}=\frac{1}{f_1}-\frac{1}{f_2}\); if \(f_1>f_2\) the result is negative, meaning diverging.

Answer 1

The Correct Option is A

Step 1: Work directly with powers.
Power of a lens is \(P = 1/f\) with sign. Convex power \(P_1 = +\dfrac{1}{f_1}\), concave power \(P_2 = -\dfrac{1}{f_2}\).
Step 2: Add the powers for lenses in contact.
\(P = P_1 + P_2 = \dfrac{1}{f_1} - \dfrac{1}{f_2}\).
Step 3: Because \(f_1 > f_2\), the convex power \(1/f_1\) is smaller in magnitude than the concave power \(1/f_2\).
So the diverging concave lens dominates and the net power is negative, \(P < 0\).
Step 4: A negative net power is the signature of a diverging system, so the pair acts as a concave lens.
\[\boxed{\text{concave lens}}\]