The refractive index of a material of a planoconcave lens is $5 / 3$, the radius of curvature is $0.3…

The refractive index of a material of a planoconcave lens is $5 / 3$, the radius of curvature is $0.3 \mathrm{~m}$. The focal length of the lens in air is
  1. $-0.45 \mathrm{~m}$
  2. $-0.6 \mathrm{~m}$
  3. $-0.75 \mathrm{~m}$
  4. $-1.0 \mathrm{~m}$

Solution

Lens maker's formula $ \frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] $ where, $R_2=\infty, R_1=0.3 \mathrm{~m}$ $ \begin{array}{llrl} \therefore & & \frac{1}{f} & =\left(\frac{5}{3}-1\right)\left(\frac{1}{0.3}-\frac{1}{\infty}\right) \\ & \Rightarrow & \frac{1}{f} & =\frac{2}{3} \times \frac{1}{0.3} \\ \text { or } & & f & =0.45 \mathrm{~m} \end{array} $

Asked in: AP EAMCET 2008

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