Physics - Geometrical Optics Question with Solution | TestHub

Using the lens maker's formula:

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

For an equiconcave lens, $R_1=-R$ and $R_2=+R$.

Given ${\mu }_g=1.5$ and ${\mu }_l=1.75$.

$\frac{1}{f}=\left(\frac{1.5}{1.75}-1\right)\left(\frac{1}{-R}-\frac{1}{R}\right)$

$\frac{1}{f}=\left(\frac{6}{7}-1\right)\left(-\frac{2}{R}\right)$

$\frac{1}{f}=\left(-\frac{1}{7}\right)\left(-\frac{2}{R}\right)=\frac{2}{7R}$

$f=\frac{7R}{2}=+3.5R$

Since the focal length is positive, the lens acts as a convex lens of focal length $3.5R$.

Answer: convex lens of focal length (3.5) R