Question

A thin convex lens and a thin concave lens are kept in contact and are co-axial. Which of the following statements is correct for this combination of two lenses ?

• A. behaves as concave lens if $|f_{convex}|>|f_{concave}|$
• B. behaves as concave lens if $|f_{convex}| < |f_{concave}|$
• C. behaves as convex lens if $|f_{convex}|>|f_{concave}|$
• D. Focal length of the lens system will change if the positions of two lenses are interchanged

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For thin lenses in contact, the equivalent power of the combination is the algebraic sum of the individual powers. The sign of the equivalent power determines the nature of the combined lens (positive for converging/convex, negative for diverging/concave).
: Key Formula or Approach:
Equivalent power \( P = P_1 + P_2 \).
In terms of focal lengths: \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \).
For a convex lens, \( f_1 = +f_{\text{convex}} \); for a concave lens, \( f_2 = -f_{\text{concave}} \).
Step 2: Detailed Explanation:
The equivalent focal length \( F \) is given by:
\[ \frac{1}{F} = \frac{1}{f_{\text{convex}}} - \frac{1}{f_{\text{concave}}} = \frac{f_{\text{concave}} - f_{\text{convex}}}{f_{\text{convex}} \cdot f_{\text{concave}}} \]
The combination behaves as a concave lens if the equivalent focal length \( F \) is negative.
For \( F<0 \), we must have:
\[ f_{\text{concave}} - f_{\text{convex}}<0 \]
\[ f_{\text{concave}}<f_{\text{convex}} \]
Taking magnitudes: \( |f_{\text{convex}}|>|f_{\text{concave}}| \).
Conversely, if \( |f_{\text{convex}}|<|f_{\text{concave}}| \), the power of the convex lens is higher, and the system behaves as a convex lens.
Interchanging the positions of thin lenses in contact does not change the equivalent focal length.
Step 3: Final Answer:
The combination behaves as a concave lens if \( |f_{\text{convex}}|>|f_{\text{concave}}| \).