Q141 / 160 · TS EAMCET 2024 (10 May Shift 2) · Physics
The power of a thin convex lens placed in air is +4 D . The refractive index of the material of the convex lens is $\frac{3}{2}$. If this convex lens is immersed in a liquid of refractive index $\frac{5}{3}$, then
- Ait behaves like a convex lens of focal length 75 cm
- Bit behaves like a convex lens of focal length 125 cm
- Correct answerCit behaves like a concave lens of focal length 125 cm
- Dit behaves like a concave lens of focal length 75 cm
Solution
For thin convex lens, $P=+4 D, \mu_{\ell}=\frac{3}{2}, \mu_m=\frac{5}{3}$
$\therefore P=\frac{1}{f_a} \Rightarrow f_a=\frac{1}{4} m=25 \mathrm{~cm}$
In air, $\frac{1}{f_a}=\left(\mu_{\ell}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$...(i) In medium, $\frac{1}{\mathrm{f}_{\mathrm{m}}}=\left(\frac{\mu_{\ell}}{\mu_{\mathrm{m}}}-1\right)\left(\frac{1}{\mathrm{R}_{\mathrm{I}}}-\frac{1}{\mathrm{R}_2}\right)$.....(ii) Dividing eq (ii) by (i) $\frac{f_m}{f_a}=\frac{\left(\mu_{\ell}-1\right)}{\left(\frac{\mu_{\ell}}{\mu_m}-1\right)}=\frac{\left(\frac{3}{2}-1\right)}{\left(\frac{3 / 2}{5 / 3}-1\right)}=-5$ $\therefore \mathrm{f}_{\mathrm{m}}=-5 \times \mathrm{f}_{\mathrm{a}}=-5 \times 25=-125 \mathrm{~cm}$
$\therefore$ The lens behaves like a concave lens of focal length 125 cm.
In air, $\frac{1}{f_a}=\left(\mu_{\ell}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$...(i) In medium, $\frac{1}{\mathrm{f}_{\mathrm{m}}}=\left(\frac{\mu_{\ell}}{\mu_{\mathrm{m}}}-1\right)\left(\frac{1}{\mathrm{R}_{\mathrm{I}}}-\frac{1}{\mathrm{R}_2}\right)$.....(ii) Dividing eq (ii) by (i) $\frac{f_m}{f_a}=\frac{\left(\mu_{\ell}-1\right)}{\left(\frac{\mu_{\ell}}{\mu_m}-1\right)}=\frac{\left(\frac{3}{2}-1\right)}{\left(\frac{3 / 2}{5 / 3}-1\right)}=-5$ $\therefore \mathrm{f}_{\mathrm{m}}=-5 \times \mathrm{f}_{\mathrm{a}}=-5 \times 25=-125 \mathrm{~cm}$
$\therefore$ The lens behaves like a concave lens of focal length 125 cm.
Hence, the correct answer is option: 'C'.