Question:
A convex lens of focal length $20 \text{ cm}$ is placed in contact with a concave lens of focal length $40 \text{ cm}$. The power of the combination is}
A convex lens of focal length $20 \text{ cm}$ is placed in contact with a concave lens of focal length $40 \text{ cm}$. The power of the combination is}
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Remember that when combining lens powers, you simply add them together. The sign of each individual power indicates whether the lens is convex or concave.
Updated On: May 31, 2026
- $+2.5 \text{ D}$
- $-2.5 \text{ D}$
- $+5 \text{ D}$
- $-5 \text{ D}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The power of a lens is defined as the reciprocal of its focal length in meters. For a combination of lenses, the total power is the sum of the individual powers of the lenses. The sign of the power indicates whether the lens is convex (positive) or concave (negative).
Step 2: Meaning
In this problem, we need to find the combined power of a convex lens and a concave lens placed in contact with each other. The focal length of the convex lens is $20 \text{ cm}$, which means its power is $\frac{1}{20} \text{ m}^{-1}$. The focal length of the concave lens is $40 \text{ cm}$, meaning its power is $-\frac{1}{40} \text{ m}^{-1}$.
Step 3: Analysis
First, convert the focal lengths to meters: \[f_1 = 20 \text{ cm} = 0.2 \text{ m}\] \[f_2 = -40 \text{ cm} = -0.4 \text{ m}\] The power of a lens is given by $P = \frac{1}{f}$, where $f$ is the focal length in meters. For the convex lens: \[P_1 = \frac{1}{0.2} = 5 \text{ D}\] For the concave lens: \[P_2 = \frac{1}{-0.4} = -2.5 \text{ D}\] The total power of the combination is the sum of the individual powers: \[P_{\text{total}} = P_1 + P_2 = 5 \text{ D} + (-2.5 \text{ D}) = 2.5 \text{ D}\]
Step 4: Conclusion
The combined power of the convex and concave lenses is $+2.5 \text{ D}$. Final Answer: (D)
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