Q144 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Physics

The focal length of biconvex lens made of glass, of equal radii is $f$. If the lens is dipped in the water, then the focal length becomes (Take, refractive index of glass and water as $\frac{3}{2}$ and $\frac{4}{3}$, respectively)

  • A
    2f
  • Correct answer
    B
    4f
  • C
    $(5 / 3) f$
  • D
    $(7 / 4) f$

Solution

Focal length of the biconvex lens is $f$. Radii are equal, i.e. $R_1=R_2=R$. $$ { }_a \mu_g=\frac{3}{2} $$ $$ { }_a \mu_w=\frac{4}{3} $$ According to lens Maker's formula, $$ \therefore \quad \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) $$ Here, $R_1=R$ and $R_2=-R$. $$ \begin{aligned} \therefore & & \frac{1}{f} & =\left(\frac{3}{2}-1\right)\left(\frac{1}{R}+\frac{1}{R}\right)=\frac{1}{2} \times \frac{2}{R}=\frac{1}{R} \\ & \Rightarrow & & f=R \end{aligned} $$ When the lens is dipped in the water, then $$ \begin{aligned} \frac{1}{f_w} & =\left(\frac{{ }_a \mu_g}{{ }_a \mu_w}-1\right)\left(\frac{1}{R}+\frac{1}{R}\right) \\ \frac{1}{f_w} & =\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right) \times\left(\frac{1}{R}+\frac{1}{R}\right) \\ & =\left(\frac{3}{2} \times \frac{3}{4}-1\right) \times \frac{2}{R}=\left(\frac{9}{8}-1\right) \times \frac{2}{R} \\ \Rightarrow \quad \frac{1}{f_w} & =\frac{1}{8} \times \frac{2}{R}=\frac{1}{4 R}=\frac{1}{4 f} \Rightarrow f_w=4 f \end{aligned} $$

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

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