Two planoconvex lenses each of focal length 1 | vedclass

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Question

(A) The focal length of a lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a plan

Vedclass · Accepted Answer

$6.67$

Vedclass · Answer

(A) The focal length of a lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a planoconvex lens, $R_1 = \infty$ and $R_2 = -R$. Thus, $\frac{1}{10} = (1.5 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} ight) = 0.5 	imes \frac{1}{R}$.This gives $R = 5 \, cm$.The middle water lens is biconcave with radii of curvature $R_1 = -5 \, cm$ and $R_2 = 5 \, cm$.The focal length $f_2$ of the water lens is $\frac{1}{f_2} = (\mu_w - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight) = (4/3 - 1) \left( \frac{1}{-5} - \frac{1}{5} ight) = (1/3) 	imes (-2/5) = -2/15 \, cm^{-1}$.The equivalent focal length $f_{eq}$ of the combination is $\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} = \frac{1}{10} - \frac{2}{15} + \frac{1}{10} = \frac{3 - 4 + 3}{30} = \frac{2}{30} = \frac{1}{15} \, cm^{-1}$.Thus, $f_{eq} = 15 \, cm = 0.15 \, m$.The optical power $P = \frac{1}{f_{eq} (	ext{in meters})} = \frac{1}{0.15} = 6.67 \, D$.

Two planoconvex lenses each of focal length $10 \, cm$ and refractive index $3/2$ are placed as shown. In the space left, water $(R.I = 4/3)$ is filled. The whole arrangement is in air. The optical power of the system is (in diopters):

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