In the figure,an air lens of radii of curvatu | vedclass

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(A) The lens is made of air $(\mu_l = 1)$ and is placed in a glass medium $(\mu_m = 1.5)$.Using the lens maker's formula for a lens in a medium:$\frac

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$15 \ cm$,concave

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(A) The lens is made of air $(\mu_l = 1)$ and is placed in a glass medium $(\mu_m = 1.5)$.Using the lens maker's formula for a lens in a medium:$\frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Given $R_1 = 10 \ cm$ and $R_2 = -10 \ cm$ (for a biconcave air lens shape):$\frac{1}{f} = \left( \frac{1}{1.5} - 1 \right) \left( \frac{1}{10} - \frac{1}{-10} \right)$$\frac{1}{f} = \left( \frac{2}{3} - 1 \right) \left( \frac{1}{10} + \frac{1}{10} \right)$$\frac{1}{f} = \left( -\frac{1}{3} \right) \left( \frac{2}{10} \right) = -\frac{2}{30} = -\frac{1}{15}$$f = -15 \ cm$.$A$ negative focal length indicates that the lens behaves as a concave lens.

In the figure,an air lens of radii of curvature $10 \ cm$ ($R_1 = 10 \ cm$,$R_2 = -10 \ cm$) is cut in a cylinder of glass $(\mu = 1.5)$. The focal length and the nature of the lens is:

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