A concave-convex lens of refractive index 1.5 | vedclass

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Question

(D) The system consists of a liquid lens (plano-concave) and a glass lens (concave-convex). For the liquid lens: $\mu_l = 1.3$,$R_1 = \infty$,$R_2 = -

Vedclass · Accepted Answer

$\frac{600}{11} \ cm$

Vedclass · Answer

(D) The system consists of a liquid lens (plano-concave) and a glass lens (concave-convex). For the liquid lens: $\mu_l = 1.3$,$R_1 = \infty$,$R_2 = -30 \ cm$. Using the lens maker's formula: $\frac{1}{f_l} = (\mu_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = (1.3 - 1) \left( \frac{1}{\infty} - \frac{1}{-30} ight) = 0.3 	imes \frac{1}{30} = \frac{0.3}{30} = \frac{1}{100} \ cm^{-1}$. For the glass lens: $\mu_g = 1.5$,$R_1 = -30 \ cm$,$R_2 = -20 \ cm$. Using the lens maker's formula: $\frac{1}{f_g} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = (1.5 - 1) \left( \frac{1}{-30} - \frac{1}{-20} ight) = 0.5 \left( \frac{-2 + 3}{60} ight) = 0.5 	imes \frac{1}{60} = \frac{1}{120} \ cm^{-1}$. The total focal length $f$ is given by $\frac{1}{f} = \frac{1}{f_l} + \frac{1}{f_g} = \frac{1}{100} + \frac{1}{120} = \frac{6 + 5}{600} = \frac{11}{600} \ cm^{-1}$. Therefore,$f = \frac{600}{11} \ cm$.

$A$ concave-convex lens of refractive index $1.5$ has radii of curvature of its surfaces as $30 \ cm$ and $20 \ cm$,respectively. The concave surface is facing upwards and is filled with a liquid of refractive index $1.3$. The focal length of the liquid-glass combination will be

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