Physics - Modern Physics Question with Solution | TestHub

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 $$\begin{array}{l}\frac{1}{f_l}=\left({}_{\ell }{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\ \Rightarrow -1=\left(\frac{1.5}{n}-1\right)\left(\frac{2}{R}\right)...(ii)\\ \text{ Dividing, (i) by (ii) }-5=\frac{0.5n}{1.5-n}\\ \Rightarrow -7.5+5n=0.5n\Rightarrow -7.5=-4.5n\\ \Rightarrow n=\frac{75}{45}=\frac{5}{3}\end{array}$$