Q: The effective focal length of the lens combination shown in Fig is – 60 cm. The radii of curvature of the curved surfaces of the Plano – convex lenses are 12 cm each and refractive index of the material of the lens is 1.5. The refractive index of the liquid is :
(a) 1.33
(b) 1.42
(c) 1.53
(d) 1.60
Ans: (d)
Solution:
Applying Lens Maker’s formula ,
$\displaystyle \frac{1}{f} = (\mu – 1)( \frac{1}{R_1} – \frac{1}{R_2}) $
$\displaystyle \frac{1}{f_1} = (1.5 – 1)( \frac{1}{-\infty} – \frac{1}{-12}) $
$\displaystyle \frac{1}{f_1} = \frac{1}{24}) $
$\displaystyle \frac{1}{f_2} = (\mu – 1)( \frac{1}{-12} – \frac{1}{12}) $
$\displaystyle \frac{1}{f_2} = – \frac{(\mu – 1)}{6} $
$\displaystyle \frac{1}{f_3} = (1.5 – 1)( \frac{1}{12} – \frac{1}{\infty}) $
$\displaystyle \frac{1}{f_3} = \frac{1}{24} $
$\displaystyle \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} $
$\displaystyle \frac{1}{-60} = \frac{1}{24} – \frac{(\mu – 1)}{6} + \frac{1}{24} $
$\displaystyle \frac{(\mu – 1)}{6} = \frac{1}{12} + \frac{1}{60} = \frac{6}{60}$
$\displaystyle \frac{(\mu – 1)}{6} = 0.1 $
μ = 1.6