The focal length of the combination of a convex lens (focal length \( f_{1} \)) and a concave lens (focal length \( f_{2} \)) is \( f \). For this combination to behave like a diverging lens
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( f_{1} \gt f_{2} \)
For two thin lenses in contact, the powers add. Here \( f_{1} \) and \( f_{2} \) are read as the magnitudes of the two focal lengths, as printed. With the sign convention, the convex lens has power \( +1/f_{1} \) and the concave lens has power \( -1/f_{2} \):
\( \displaystyle \frac{1}{f} = \frac{1}{f_{1}} - \frac{1}{f_{2}} \)
The combination is diverging when its net power is negative:
\( \displaystyle \frac{1}{f_{1}} - \frac{1}{f_{2}} \lt 0 \ \Rightarrow\ \frac{1}{f_{1}} \lt \frac{1}{f_{2}} \ \Rightarrow\ f_{1} \gt f_{2} \)
So the concave lens must be the more powerful of the two, i.e. it must have the shorter focal length.
Note: It seems that there is a mistake in the original NTA paper, which prints Options (3) and (4) identically as \( f_{1} \lt f_{2} \). The intended form of one of them cannot be recovered, but neither is the answer, so the question can still be solved.