KCET · Physics · Ray Optics
Radii of curvature of a converging lens are in the ratio \(1: 2\). Its focal length is \(6 \mathrm{~cm}\) and refractive index is \(1.5\). Then its radii of curvature are ......... respectively.
- A \(9 \mathrm{~cm}\) and \(18 \mathrm{~cm}\)
- B \(6 \mathrm{~cm}\) and \(12 \mathrm{~cm}\)
- C \(3 \mathrm{~cm}\) and \(6 \mathrm{~cm}\)
- D \(4.5 \mathrm{~cm}\) and \(9 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(D) \(4.5 \mathrm{~cm}\) and \(9 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
Given, \(\frac{R_{1}}{R_{2}}=\frac{1}{2}, f=6 \mathrm{~cm}\) and \(\mu=1.5\)
Let \(\quad R_{1}=R\) and \(R_{2}=2 R\)
We know the focal length of converging lens
By using relation \(\frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}+\frac{1}{R_{2}}\right]\)
Substituting all the values, we get
\(\begin{gathered}
R=4.5 \mathrm{~cm} \\
\therefore \quad R_{1}=4.5 \mathrm{~cm} \text { and } R_{2}=2 \times 4.5=9 \mathrm{~cm}
\end{gathered}\)
Let \(\quad R_{1}=R\) and \(R_{2}=2 R\)
We know the focal length of converging lens
By using relation \(\frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}+\frac{1}{R_{2}}\right]\)
Substituting all the values, we get
\(\begin{gathered}
R=4.5 \mathrm{~cm} \\
\therefore \quad R_{1}=4.5 \mathrm{~cm} \text { and } R_{2}=2 \times 4.5=9 \mathrm{~cm}
\end{gathered}\)
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