Question

A thin convex lens and a thin concave lens are kept coaxially in contact. Choose the correct option.

• A. Focal length changes when positions of lenses are interchanged
• B. Behaves as a convex lens when \(f_{\text{convex}} > f_{\text{concave}}\)
• C. Behaves as a concave lens when \(f_{\text{concave}} > f_{\text{convex}}\)
• D. Behaves as convex when \(f_{\text{convex}} < f_{\text{concave}}\)

Hint:

For lenses in contact: \[ \frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2} \] If \(f_{eq} > 0\) → convex behaviour If \(f_{eq} > 0\) → concave behaviour.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When two thin lenses are placed coaxially in contact, their equivalent power is the algebraic sum of their individual powers.
Consequently, the equivalent focal length depends on the reciprocal sum of their individual focal lengths.
Step 2: Key Formula or Approach:
The equivalent focal length $F$ for lenses in contact is given by:
\[ \frac{1}{F_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} \]
Step 3: Detailed Explanation:
Let $f_1$ be the focal length of the convex lens ($f_1 = +f_c>0$) and $f_2$ be the focal length of the concave lens ($f_2 = -f_v<0$, where $f_v$ is its positive magnitude).
\[ \frac{1}{F_{eq}} = \frac{1}{f_c} - \frac{1}{f_v} = \frac{f_v - f_c}{f_c f_v} \]
For the combination to behave as a **convex lens**, the equivalent focal length $F_{eq}$ must be positive.
\[ F_{eq}>0 \implies \frac{f_v - f_c}{f_c f_v}>0 \]
Since $f_c$ and $f_v$ are absolute magnitudes, their product $f_c f_v$ is strictly positive.
Therefore, the numerator must be positive:
\[ f_v - f_c>0 \implies f_v>f_c \]
This means the magnitude of the focal length of the concave lens ($f_v$) must be GREATER than the focal length of the convex lens ($f_c$).
In words: "f of convex<f of concave".
Let's analyze the options:
(A) Interchange of position doesn't alter the formula $1/F_{eq} = 1/f_1 + 1/f_2$, so $F_{eq}$ remains identical.
(B) If $f_c>f_v$, then $F_{eq}$ is negative, making it behave as a concave lens.
(C) If $f_v>f_c$, then $F_{eq}$ is positive, making it behave as a convex lens.
(D) If $f_c<f_v$, then $F_{eq}$ is positive, making it behave as a convex lens. This statement is accurate.
Step 4: Final Answer:
The combination behaves as a convex lens when $f_{\text{convex}}<f_{\text{concaves}}$.