MHT CET · Physics · Experimental Physics
A glass convex lens is of refractive index \(1 \cdot 55\) with both faces of same radius of curvature. What will be the radius of curvature if focal length is to be \(20 \mathrm{~cm}\) ?
- A \(22 \mathrm{~cm}\)
- B \(21 \mathrm{~cm}\)
- C \(18 \mathrm{~cm}\)
- D \(20 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(A) \(22 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Lens maker's formula,
\(\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Here, \(\mathrm{f}=20 \mathrm{~cm}, \mu=1.55, \mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=-\mathrm{R}\)
\(\frac{1}{20}=(1.55-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right) \text { or } \frac{1}{20}=\) \(0.55 \times \frac{2}{R} \)
\( \Rightarrow R=1.1 \times 20=22 \mathrm{~cm}\)
\(\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Here, \(\mathrm{f}=20 \mathrm{~cm}, \mu=1.55, \mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=-\mathrm{R}\)
\(\frac{1}{20}=(1.55-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right) \text { or } \frac{1}{20}=\) \(0.55 \times \frac{2}{R} \)
\( \Rightarrow R=1.1 \times 20=22 \mathrm{~cm}\)
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