Two identical thin plano-convex glass lenses | vedclass

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(C) Using the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For the two identical plano-convex lenses (

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$-50$

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(C) Using the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For the two identical plano-convex lenses ($f_1$ and $f_2$):$\frac{1}{f_1} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{\infty} ight) = 0.5 	imes \frac{1}{20} = \frac{1}{40} \, cm^{-1}$.Similarly,$\frac{1}{f_2} = \frac{1}{40} \, cm^{-1}$.The oil lens formed in the middle is a biconcave lens with radii of curvature $R_1 = -20 \, cm$ and $R_2 = 20 \, cm$ (or vice versa,resulting in the same focal length).For the oil lens $(f_3)$:$\frac{1}{f_3} = (1.7 - 1) \left( \frac{1}{-20} - \frac{1}{20} ight) = 0.7 	imes \left( -\frac{2}{20} ight) = 0.7 	imes \left( -\frac{1}{10} ight) = -\frac{0.7}{10} = -\frac{7}{100} \, cm^{-1}$.The focal length of the combination is given by $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$.$\frac{1}{f} = \frac{1}{40} + \frac{1}{40} - \frac{7}{100} = \frac{2}{40} - \frac{7}{100} = \frac{1}{20} - \frac{7}{100}$.$\frac{1}{f} = \frac{5 - 7}{100} = -\frac{2}{100} = -\frac{1}{50}$.Therefore,$f = -50 \, cm$.

Two identical thin plano-convex glass lenses (refractive index $1.5$) each having a radius of curvature of $20\, cm$ are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index $1.7$. The focal length of the combination is.......$cm$.

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