# A concave lens of glass, refractive index 1.5 both surface of same radius of curvature R. On immersion in a medium of refractive index 1.75 it will behave as lens…

Q: A concave lens of glass, refractive index 1.5 both surface of same radius of curvature R. On immersion in a medium of refractive index 1.75 it will behave as lens…

Sol: When glass lens is immersed in a medium, its refractive index is mμg

$\large ^m\mu_g = \frac{\mu_g}{\mu_m} = \frac{1.5}{1.75}$

$\large ^m\mu_g = \frac{6}{7}$

By lens makers’s formula

$\large \frac{1}{f} = (^m\mu_g -1)(\frac{1}{R_1} -\frac{1}{R_2})$

$\large \frac{1}{f} = (\frac{6}{7} -1)(-\frac{1}{R} -\frac{1}{R})$

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

f = 7R/2 = 3.5 R

Hence, the given lens in medium behaves like convergent lens of focal length 3.5 R