Physics - Geometrical Optics Question with Solution | TestHub

Using the lens maker's formula in air:

$P_a=({\mu }_g-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$3=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{3}{0.5}=6$

When the lens is placed in a liquid, the new power $P_l$ is given by:

$P_l=\left(\frac{{\mu }_g}{{\mu }_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Substituting the given values ${\mu }_g=1.5$ and ${\mu }_l=2$:

$P_l=\left(\frac{1.5}{2}-1\right)\times 6$

$P_l=(0.75-1)\times 6$

$P_l=-0.25\times 6=-1.5\mathrm{D}$

Answer: $-1.5\mathrm{D}$