Two identical thin biconvex lenses of focal l | vedclass

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(A) For a biconvex lens,the lens maker's formula is $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Given $f = 15 \, cm$ and $\

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$10$

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(A) For a biconvex lens,the lens maker's formula is $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Given $f = 15 \, cm$ and $\mu = 1.5$,and for a symmetric biconvex lens $R_1 = R$ and $R_2 = -R$,we have:$\frac{1}{15} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = 0.5 \left( \frac{2}{R} \right) = \frac{1}{R}$.Thus,$R = 15 \, cm$.The combination consists of two biconvex lenses and a liquid lens in between. The liquid lens has refractive index $\mu_l = 1.25$ and its surfaces have radii of curvature $-R$ and $R$ (concave shape).The focal length of the liquid lens $f_l$ is given by:$\frac{1}{f_l} = (\mu_l - 1) \left( \frac{1}{-R} - \frac{1}{R} \right) = (1.25 - 1) \left( -\frac{2}{R} \right) = 0.25 \left( -\frac{2}{15} \right) = -\frac{0.5}{15} = -\frac{1}{30}$.The equivalent focal length $f_{eq}$ of the combination is:$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_l} + \frac{1}{f_2} = \frac{1}{15} - \frac{1}{30} + \frac{1}{15} = \frac{2 - 1 + 2}{30} = \frac{3}{30} = \frac{1}{10}$.Therefore,$f_{eq} = 10 \, cm$.

Two identical thin biconvex lenses of focal length $15 \, cm$ and refractive index $1.5$ are in contact with each other. The space between the lenses is filled with a liquid of refractive index $1.25$. The focal length of the combination is $cm$.

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