Question

A composite lens is made by filling the two halves of the lens by medium of refractive index $\mu_1=1.5$ and $\mu_2=2$ as shown. The radius of curvature of both sides of lens is $R_1$ and $R_2$ of 30 cm and 45 cm respectively. The refractive index of a liquid medium in which if this lens is dipped completely, the effective power of the lens becomes zero is $(1+\frac{x}{10})$, then x is

• A. 5
• B. -5
• C. 10
• D. -10

Answer 1

Answer: (B)

-5

Answer 2

The composite lens can be treated as two separate thin lenses placed in contact. The left half is a plano-convex lens with refractive index $\mu_1 = 1.5$ and radius of curvature $r_{11} = +30$ cm. The right half is a plano-convex lens with refractive index $\mu_2 = 2$ and radius of curvature $r_{22} = -45$ cm (assuming the lens is oriented such that the curvature is negative from the perspective of the light path).

The power of a thin lens when immersed in a medium of refractive index $\mu_L$ is given by the formula: $P = (\frac{\mu_{lens}}{\mu_L} - 1) (\frac{1}{r_1} - \frac{1}{r_2})$

For the left half (lens 1): $\mu_1 = 1.5$ $r_{11} = +30$ cm (convex surface) $r_{12} = \infty$ (plane surface) $P_1 = \left(\frac{1.5}{\mu_L} - 1\right) \left(\frac{1}{30} - \frac{1}{\infty}\right) = \left(\frac{1.5}{\mu_L} - 1\right) \frac{1}{30}$

For the right half (lens 2): $\mu_2 = 2$ $r_{21} = \infty$ (plane surface) $r_{22} = -45$ cm (convex surface, but the formula uses $1/r_2$, and the curvature is towards the left, so its contribution to power is negative when using the standard lens maker formula convention. However, the provided solution uses $1/45$ with a negative sign in the power formula, effectively $1/r_2 = -1/45$. Let's follow the solution's convention for consistency.) $P_2 = \left(\frac{2}{\mu_L} - 1\right) \left(\frac{1}{\infty} - \frac{1}{45}\right) = \left(\frac{2}{\mu_L} - 1\right) \left(-\frac{1}{45}\right)$

The effective power of the composite lens is the sum of the powers of its two halves: $P_{total} = P_1 + P_2$. We are given that the effective power becomes zero: $P_{total} = 0$. $P_1 + P_2 = 0$ $\left(\frac{1.5}{\mu_L} - 1\right) \frac{1}{30} + \left(\frac{2}{\mu_L} - 1\right) \left(-\frac{1}{45}\right) = 0$ Multiply by 90 to clear denominators: $3 \left(\frac{1.5}{\mu_L} - 1\right) - 2 \left(\frac{2}{\mu_L} - 1\right) = 0$ $\frac{4.5}{\mu_L} - 3 - \frac{4}{\mu_L} + 2 = 0$ $\frac{0.5}{\mu_L} - 1 = 0$ $\frac{0.5}{\mu_L} = 1$ $\mu_L = 0.5$

The problem states that the refractive index of the liquid is $(1 + \frac{x}{10})$. So, $\mu_L = 1 + \frac{x}{10}$. $0.5 = 1 + \frac{x}{10}$ $0.5 - 1 = \frac{x}{10}$ $-0.5 = \frac{x}{10}$ $x = -0.5 \times 10 = -5$.