Question:
Two lenses of powers \(-1.75 \, \text{D}\) and \(+2.25 \, \text{D}\) are placed in contact. Find the focal length of the combination.
Two lenses of powers \(-1.75 \, \text{D}\) and \(+2.25 \, \text{D}\) are placed in contact. Find the focal length of the combination.
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For lenses in contact, total power \(P = P_1 + P_2\) and focal length \(f = 1/P\) (in meters). Convert to desired units as needed.
Updated On: Jul 18, 2026
- 100 cm
- 50 cm
- 200 cm
- 150 cm
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The Correct Option is C
Solution and Explanation
Step 1: Recall relation between power and focal length.
Power of lens \(P = \frac{100}{f (\text{cm})}\) or \(P (\text{D}) = \frac{1}{f (\text{m})}\).
Step 2: Total power of lenses in contact.
\[ P_{\text{total}} = P_1 + P_2 = -1.75 + 2.25 = 0.50 \, \text{D} \]
Step 3: Relation between total power and focal length.
\[ f_{\text{total}} = \frac{1}{P_{\text{total}}} = \frac{1}{0.50} = 2 \, \text{m} \]
Step 4: Convert to cm.
\[ f_{\text{total}} = 2 \times 100 = 200 \, \text{cm} \]
Step 5: Verify reasoning.
Positive total power indicates converging combination. Focal length consistent with lens powers.
Step 6: Final conclusion.
Hence, the focal length of the combination is:
\[ \boxed{200 \, \text{cm}} \]
Power of lens \(P = \frac{100}{f (\text{cm})}\) or \(P (\text{D}) = \frac{1}{f (\text{m})}\).
Step 2: Total power of lenses in contact.
\[ P_{\text{total}} = P_1 + P_2 = -1.75 + 2.25 = 0.50 \, \text{D} \]
Step 3: Relation between total power and focal length.
\[ f_{\text{total}} = \frac{1}{P_{\text{total}}} = \frac{1}{0.50} = 2 \, \text{m} \]
Step 4: Convert to cm.
\[ f_{\text{total}} = 2 \times 100 = 200 \, \text{cm} \]
Step 5: Verify reasoning.
Positive total power indicates converging combination. Focal length consistent with lens powers.
Step 6: Final conclusion.
Hence, the focal length of the combination is:
\[ \boxed{200 \, \text{cm}} \]
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