The refractive index of a material of a plano | vedclass

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(A) For a plano-concave lens, one surface is plane $(R_1 = \infty)$ and the other is concave $(R_2 = 0.3 \,m)$. By convention, for a concave surface,

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$-0.45$

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(A) For a plano-concave lens, one surface is plane $(R_1 = \infty)$ and the other is concave $(R_2 = 0.3 \,m)$. By convention, for a concave surface, $R_2 = -0.3 \,m$.Using the lens maker's formula:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$Given $\mu = 5/3$, $R_1 = \infty$, and $R_2 = -0.3 \,m$:$\frac{1}{f} = \left( \frac{5}{3} - 1 ight) \left( \frac{1}{\infty} - \frac{1}{-0.3} ight)$$\frac{1}{f} = \left( \frac{2}{3} ight) \left( 0 + \frac{1}{0.3} ight)$$\frac{1}{f} = \frac{2}{3} 	imes \frac{10}{3} = \frac{20}{9}$$f = \frac{9}{20} = 0.45 \,m$Since it is a concave lens, the focal length must be negative by sign convention, so $f = -0.45 \,m$.

The refractive index of a material of a plano-concave lens is $5/3$, and the radius of curvature is $0.3 \,m$. The focal length of the lens in air is (in $\,m$)

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