Two planoconvex lenses each of focal length 1 | vedclass

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(A) For the two planoconvex lenses, $f_1 = 10 \, cm$ and $f_2 = 10 \, cm$. Using the lens maker's formula $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1

Vedclass · Accepted Answer

$6.67$

Vedclass · Answer

(A) For the two planoconvex lenses, $f_1 = 10 \, cm$ and $f_2 = 10 \, cm$. Using the lens maker's formula $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$:$\frac{1}{10} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{\infty} - \frac{1}{-R} \right) \implies \frac{1}{10} = \frac{1}{2} \cdot \frac{1}{R} \implies R = 5 \, cm$.The water lens formed in the middle is a biconcave lens with radii of curvature $R_1 = -5 \, cm$ and $R_2 = 5 \, cm$. Its focal length $f_3$ is given by:$\frac{1}{f_3} = \left( \frac{4}{3} - 1 \right) \left( \frac{1}{-5} - \frac{1}{5} \right) = \left( \frac{1}{3} \right) \left( -\frac{2}{5} \right) = -\frac{2}{15} \, cm^{-1}$.The effective focal length $f_{eff}$ of the combination is:$\frac{1}{f_{eff}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} = \frac{1}{10} + \frac{1}{10} - \frac{2}{15} = \frac{3 + 3 - 4}{30} = \frac{2}{30} = \frac{1}{15} \, cm^{-1}$.Thus, $f_{eff} = 15 \, cm = 0.15 \, m$.The optical power $P$ in diopters is:$P = \frac{1}{f_{eff} (in \, meters)} = \frac{1}{0.15} = \frac{100}{15} \approx 6.67 \, D$.

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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