Ray Optics and Optical Instruments Part 3

Concept:The power of a lens depends on its refractive index and radii of curvature as given by Lens Maker's formula.Given that the magnitudes of power of two lenses are equal and one surface is shared, we can equate their absolute powers to find the ratio of radii.Formula:Lens Maker's formula: $P = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$where $P$ is power, $\mu$ is refractive index, $R_1$ and $R_2$ are radii of curvature with sign convention.Solution:Let $R_1$ and $R_2$ be the front and back surface radii of the biconvex lens (both positive for convex surfaces).For biconvex lens ($\mu = 1.5$): $P_{\text{convex}} = (1.5-1)\left(\frac{1}{R_1} - \frac{1}{-R_2}\right) = 0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$.For plano-concave lens ($\mu = 1.7$), one surface is flat ($R_3 = \infty$), the curved surface is concave with radius $R_4 = R_2$ (since it matches back surface of biconvex).Using sign convention: $R_4 = -R_2$ for concave surface? Wait, careful: For a plano-concave lens, the concave surface has its center of curvature on the same side as the incoming light. Usually, if the flat side is on the left, the concave surface is on the right, so $R_4$ is negative. But in the existing solution they used $R_4 = R_2$ with a negative sign in formula? Let's re-derive cleanly: For a plano-concave lens, one surface is plane ($R = \infty$), the other is concave (radius negative). If the curved surface radius is $R_2$ (magnitude), then $R_4 = -R_2$. The formula: $P_{\text{concave}} = (1.7-1)\left(\frac{1}{\infty} - \frac{1}{-R_2}\right) = 0.7\left(0 + \frac{1}{R_2}\right) = \frac{0.7}{R_2}$? That gives positive power, but a concave lens should have negative power. The sign convention: For a concave surface, if the light comes from left, the radius of curvature is negative if the center is on the left. In the existing solution they got $P_{\text{concave}} = -\frac{0.7}{R_2}$, which is correct. So they used $R_4 = R_2$ (positive) in the formula with $-1/R_4$? Let's check: They wrote $(1.7-1)\left(\frac{1}{R_3}-\frac{1}{R_4}\right)$ with $R_3=\infty$ and $R_4=R_2$. But that yields $0.7(0 - 1/R_2) = -0.7/R_2$. So they kept $R_4$ as positive but the formula includes a minus sign, so the concave surface gives negative contribution. This is consistent with standard convention: $R$ is positive if the center of curvature is on the opposite side of the incident light. For a plano-concave lens, if the curved surface is on the right, the incident light from left sees that surface as concave, so $R$ is negative. But using the formula with $R_4 = +R_2$ and the minus sign inside the bracket yields the same result. So we can present it as: $P_{\text{concave}} = (1.7-1)\left(\frac{1}{\infty} - \frac{1}{R_2}\right) = 0.7\left(-\frac{1}{R_2}\right) = -\frac{0.7}{R_2}$. Keep it simple.Magnitudes are equal: $\left|P_{\text{convex}}\right| = \left|P_{\text{concave}}\right|$So $0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{0.7}{R_2}$ (ignoring negative sign as magnitude).Multiply both sides by $R_2$: $0.5\left(\frac{R_2}{R_1} + 1\right) = 0.7$Divide by $0.5$: $\frac{R_2}{R_1} + 1 = \frac{0.7}{0.5} = 1.4$So $\frac{R_2}{R_1} = 0.4 = \frac{2}{5}$Thus $\frac{R_1}{R_2} = \frac{5}{2}$.Answer:The ratio $R_1 : R_2 = 5 : 2$, which corresponds to option (B).