Q142 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Physics

A convex lens of focal length 25 cm and made of glass with refractive index 1.5 is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water = $\frac{4}{3}$ ]

  • A
    100 cm
  • B
    37.5 cm
  • Correct answer
    C
    75 cm
  • D
    12.5 cm

Solution

We have $$ \begin{aligned} & \frac{1}{\mathrm{f}_{\text {water }}}=\left({ }^{\mathrm{w}} \mu_{\mathrm{g}}-1\right)\left(\frac{2}{\mathrm{R}}\right) \\ & \text { and, } \frac{1}{\mathrm{f}_{\mathrm{a}}}=\left({ }^{\mathrm{a}} \mu_{\mathrm{g}}-1\right)\left(\frac{2}{\mathrm{R}}\right) \\ & \text { So, } \frac{\mathrm{f}_{\mathrm{a}}}{\mathrm{f}_{\mathrm{w}}}=\frac{\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1}{\mu_{\mathrm{g}}-1}=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4} \\ & \Rightarrow \mathrm{f}_{\mathrm{w}}=4 \mathrm{f}_{\mathrm{a}}=100 \mathrm{~cm} \\ & \text { Now, } \Delta \mathrm{f}=(100-25) \mathrm{cm}=75 \mathrm{~cm} \end{aligned} $$

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

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