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
(A) convergent lens of focal length 3.5 R
(B) convergent lens of focal length 3.0 R
(C) divergent lens of focal length 3.5 R
(D) divergent lens of focal length 3.0 R
Ans: (A)
Sol: Using Lens maker’s formula
$\large \frac{1}{f} = (\frac{\mu_g}{\mu_m}-1)(\frac{1}{R_1} – \frac{1}{R_2})$
$\large \frac{1}{f} = (\frac{1.5}{1.75}-1)(\frac{1}{-R} – \frac{1}{R}) = \frac{1}{3.5R}$
f = 3.5 R