Q125 / 180 · KCET 2013 · Physics

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 $6 \mathrm{~cm}$
  • Correct answer
    D
    $4.5 \mathrm{~cm}$ and $9 \mathrm{~cm}$

Solution

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 $\begin{gathered} R=4.5 \mathrm{~cm} \\ \therefore \quad R_{1}=4.5 \mathrm{~cm} \text { and } R_{2}=2 \times 4.5=9 \mathrm{~cm} \end{gathered}$

Hence, the correct answer is option: 'D'.

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