Q105 / 120 · VITEEE 2010 · Physics

The power of a thin convex lens $\left(\mathrm{a}_{\mathrm{g}}=1.5\right)$ is $5.0 \mathrm{D} .$ When it is placed in a liquid of refractive index $_{a} n_{b}$, then it behaves as a concave lens of focal length $100 \mathrm{~cm}$. The refractive index of the liquid $\mathrm{a}_{l}$ will be

  • Correct answer
    A
    $5 \sqrt{3}$
  • B
    $4 / 3$
  • C
    $\sqrt{3}$
  • D
    $5 / 4$

Solution

By using lens maker's formula, $$ \begin{array}{l} \frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ \Rightarrow 5=(1.5-1)\left(\frac{2}{R}\right)...(i) \end{array} $$ If a lens of refractive index $\mu_{g}$ is immersed in a liquid of refractive index $\mu_{1}$, then its focal length in liquid $$ \begin{array}{l} \frac{1}{f_{l}}=\left(_{\ell} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ \Rightarrow-1=\left(\frac{1.5}{n}-1\right)\left(\frac{2}{R}\right) ...(ii)\\ \text { Dividing, (i) by (ii) }-5=\frac{0.5 n}{1.5-n} \\ \Rightarrow-7.5+5 n=0.5 n \Rightarrow-7.5=-4.5 n \\ \Rightarrow n=\frac{75}{45}=\frac{5}{3} \end{array} $$

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

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