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GUJCET · Physics · Ray Optics and Optical Instruments
Double convex lenses are to be manufactured from a glass of refractive index 1.55 with both bases of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm ?
- A 44 cm
- B 2.2 cm
- C 22 cm
- D 4.4 cm
Answer & Solution
Correct Answer
(C) 22 cm
Step-by-step Solution
Detailed explanation
(C) 22 cm
\(f=+20 \mathrm{~cm} \quad n=1.55\)
\(\mathbf{R}_{1}=+\mathbf{R} \quad \mathbf{R}_{2}=-\mathbf{R}\)
from lens maker's formula,
\(
\begin{aligned}
& \frac{1}{f}=(n-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right) \\
\therefore \quad & \frac{1}{20}=(1.55-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{-\mathrm{R}}\right) \\
\therefore \quad & \frac{1}{20}=0.55 \times \frac{2}{\mathrm{R}} \\
\therefore & \mathrm{R}=20 \times 0.55 \times 2 \\
\therefore & \mathrm{R}=22 \mathrm{~cm}
\end{aligned}
\)
\(f=+20 \mathrm{~cm} \quad n=1.55\)
\(\mathbf{R}_{1}=+\mathbf{R} \quad \mathbf{R}_{2}=-\mathbf{R}\)
from lens maker's formula,
\(
\begin{aligned}
& \frac{1}{f}=(n-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right) \\
\therefore \quad & \frac{1}{20}=(1.55-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{-\mathrm{R}}\right) \\
\therefore \quad & \frac{1}{20}=0.55 \times \frac{2}{\mathrm{R}} \\
\therefore & \mathrm{R}=20 \times 0.55 \times 2 \\
\therefore & \mathrm{R}=22 \mathrm{~cm}
\end{aligned}
\)
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