# The effective focal length of the lens combination shown in Fig is – 60 cm. The radii of curvature of the curved surfaces …

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