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