MHT CET · Physics · Ray Optics
A glass convex lens is of refractive index 1.55 with both faces of same radius of curvature. What is the radius of curvature required if focal length is to be \(20 \mathrm{~cm}\) ?
- A \(21 \mathrm{~cm}\)
- B \(18 \mathrm{~cm}\)
- C \(20 \mathrm{~cm}\)
- D \(22 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(D) \(22 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Using the Lens maker's formula,
\(\begin{aligned} & f_1=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ & \text { Here, } f=20 \mathrm{~cm}, \mu 1.55, R_1=R, R_2=-R \\ & \Rightarrow \frac{1}{20}=(1.55-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right) \\ & \Rightarrow \frac{1}{20}=0.55 \times \frac{2}{R} \\ & \Rightarrow R=1.1 \times 20=22 \mathrm{~cm}\end{aligned}\)
\(\begin{aligned} & f_1=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ & \text { Here, } f=20 \mathrm{~cm}, \mu 1.55, R_1=R, R_2=-R \\ & \Rightarrow \frac{1}{20}=(1.55-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right) \\ & \Rightarrow \frac{1}{20}=0.55 \times \frac{2}{R} \\ & \Rightarrow R=1.1 \times 20=22 \mathrm{~cm}\end{aligned}\)
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