Q. A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a
IIT JEEIIT JEE 1999
Solution:
$R_1=-R,R_2+R, \mu_g=1.5 and \mu_m=1.75$
$\therefore \, \, \, \, \, \, \, \, \, \, \frac{1}{f}=\Bigg(\frac{\mu_g}{\mu_m}-1\Bigg)\Bigg(\frac{1}{R_1}-\frac{1}{R_2}\Bigg)$
Substituting the values, we have
$\frac{1}{f}=\Bigg(\frac{1.5}{1.75}-1\Bigg)\Bigg(\frac{1}{-R}-\frac{1}{R}\Bigg)=\frac{1}{3.5 R}$
$\therefore \, \, \, \, \, \, \, \, \, \, \, f=+3.5 R$
Therefore, in the medium it will behave like a convergent
lens of focal length 3.5R. It can be understood as, $\mu_m > \mu_g,$
the lens will change its behaviour.
