The two surfaces of a concave lens, made of g | vedclass

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(A) Using the lens maker's formula:$\frac{1}{f} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Given that the r

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becomes a convergent lens of focal length $3.5 R$

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(A) Using the lens maker's formula:$\frac{1}{f} = \left( \frac{\mu_g}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Given that the refractive index of the glass lens is $\mu_g = 1.5$ and the refractive index of the surrounding medium is $\mu_m = 1.75$.For a biconcave lens, the radii of curvature are $R_1 = -R$ and $R_2 = +R$.Substituting these values into the formula:$\frac{1}{f} = \left( \frac{1.5}{1.75} - 1 \right) \left( \frac{1}{-R} - \frac{1}{R} \right)$$\frac{1}{f} = \left( \frac{1.5 - 1.75}{1.75} \right) \left( -\frac{2}{R} \right)$$\frac{1}{f} = \left( \frac{-0.25}{1.75} \right) \left( -\frac{2}{R} \right)$$\frac{1}{f} = \left( -\frac{1}{7} \right) \left( -\frac{2}{R} \right) = \frac{2}{7R}$$f = +3.5 R$Since the focal length $f$ is positive, the lens behaves as a convergent lens.

The two surfaces of a concave lens, made of glass of refractive index $1.5$, have the same radii of curvature $R$. It is now immersed in a medium of refractive index $1.75$. Then the lens:

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