Q123 / 180 · KCET 2022 · Physics

The power of a equi-concave lens is $-4.5 \mathrm{D}$ and is made of a material of refractive index 1.6, the radii of curvature of the lens is

  • A
    $36.6 \mathrm{~cm}$
  • B
    $-266 \mathrm{~cm}$
  • C
    $115.44 \mathrm{~cm}$
  • Correct answer
    D
    $-26.6 \mathrm{~cm}$

Solution

Power of equi-concave lens, $P=-4.5 \mathrm{D}$ Refractive index, $\mu=1.6$ Focal length of lens, $$ f=\frac{1}{P}=\left(\frac{1}{-4.5}\right) \mathrm{m}=\left(\frac{100}{-4.5}\right) \mathrm{cm}=\left(\frac{-200}{9}\right) \mathrm{cm} $$ By lens Maker's formula, we get $$ \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) $$ For equi-concave lens, $$ R_1=-R \text { and } R_2=R $$ Putting these values in Eq. (i), we get $$ \begin{aligned} & \frac{1}{200 / 9}=(1.6-1)\left(\frac{1}{-R}-\frac{1}{R}\right) \Rightarrow \frac{9}{200}=0.6\left(\frac{-2}{R}\right) \\ & \Rightarrow \quad R=\frac{-0.6 \times 2 \times 200}{9}=-26.6 \mathrm{~cm} \end{aligned} $$

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

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