Question:
If the radius of curvature of the curved surface of a plano-convex lens is $50$ cm, its focal length is ($\mu = 1.5$)
If the radius of curvature of the curved surface of a plano-convex lens is $50$ cm, its focal length is ($\mu = 1.5$)
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Plano-convex: use only curved surface in formula.
Updated On: May 2, 2026
- $0.5$ m
- $0.75$ m
- $1.25$ m
- $0.25$ m
- $1$ m
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Solution and Explanation
Concept: Lens maker formula
\[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] ---
Step 1: Plano-convex lens
• One surface plane → $R_2 = \infty$
• One curved → $R_1 = 50$ cm = 0.5 m ---
Step 2: Substitute \[ \frac{1}{f} = (1.5 - 1)\left(\frac{1}{0.5} - 0\right) \] ---
Step 3: Simplify \[ \frac{1}{f} = 0.5 \times 2 = 1 \] \[ f = 1 \text{ m} \] --- Physical Insight:
• Higher refractive index → smaller focal length
• Plano surface contributes nothing to curvature --- Final Answer: \[ \boxed{1 \text{ m}} \]
\[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] ---
Step 1: Plano-convex lens
• One surface plane → $R_2 = \infty$
• One curved → $R_1 = 50$ cm = 0.5 m ---
Step 2: Substitute \[ \frac{1}{f} = (1.5 - 1)\left(\frac{1}{0.5} - 0\right) \] ---
Step 3: Simplify \[ \frac{1}{f} = 0.5 \times 2 = 1 \] \[ f = 1 \text{ m} \] --- Physical Insight:
• Higher refractive index → smaller focal length
• Plano surface contributes nothing to curvature --- Final Answer: \[ \boxed{1 \text{ m}} \]
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