Two plano-convex lenses,each of focal length | vedclass

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(A) $1$. For a plano-convex lens,the focal length $f_1$ is given by the lens maker's formula: $\frac{1}{f_1} = (\mu_g - 1) \left( \frac{1}{R_1} - \fra

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$6.67$

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(A) $1$. For a plano-convex lens,the focal length $f_1$ is given by the lens maker's formula: $\frac{1}{f_1} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Given $f_1 = 10 \, cm$ and $\mu_g = 3/2$. For a plano-convex lens,$R_1 = \infty$ and $R_2 = -R$.$\frac{1}{10} = (\frac{3}{2} - 1) \left( \frac{1}{\infty} - \frac{1}{-R} ight) = \frac{1}{2} \cdot \frac{1}{R} \Rightarrow R = 5 \, cm$.$2$. The water lens formed between the two plano-convex lenses is a biconcave (equi-concave) lens with radii of curvature $R_1 = -5 \, cm$ and $R_2 = 5 \, cm$.The focal length $f_w$ of this water lens is: $\frac{1}{f_w} = (\mu_w - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = (\frac{4}{3} - 1) \left( \frac{1}{-5} - \frac{1}{5} ight) = \frac{1}{3} \left( -\frac{2}{5} ight) = -\frac{2}{15} \, cm^{-1}$.So,$f_w = -7.5 \, cm$.$3$. The system consists of three lenses in contact: two plano-convex lenses and one biconcave water lens.The net focal length $f_{	ext{net}}$ is given by: $\frac{1}{f_{	ext{net}}} = \frac{1}{f_1} + \frac{1}{f_w} + \frac{1}{f_2} = \frac{1}{10} - \frac{2}{15} + \frac{1}{10} = \frac{3 - 4 + 3}{30} = \frac{2}{30} = \frac{1}{15} \, cm^{-1}$.Thus,$f_{	ext{net}} = 15 \, cm = 0.15 \, m$.$4$. The optical power $P$ is $P = \frac{1}{f_{	ext{net}} (	ext{in meters})} = \frac{1}{0.15} = 6.67 \, D$.

Two plano-convex lenses,each of focal length $10 \, cm$ and refractive index $3/2$,are placed as shown in the figure. The space between them is filled with water $\left( \mu = 4/3 \right)$. The whole arrangement is in air. The optical power of the system is ....... $D$.

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