Question

The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then the refractive index of the material of the lens is

• A. greater than zero but less than 1.5
• B. greater than 1.5 but less than 2.0
• C. greater than 2.0 but less than 2.5
• D. greater than 2.5 but less than 2.0

Answer 1

Answer: (A)

greater than zero but less than 1.5

Answer 2

An equi-convex lens has two convex surfaces with the same radius of curvature. Let the magnitude of the radius of curvature be $R$. For the first surface encountered by light, the radius of curvature is $R_1 = +R$. For the second surface, the radius of curvature is $R_2 = -R$.

The lens maker's formula relates the focal length $f$, the refractive index of the lens material $\mu$, and the radii of curvature $R_1$ and $R_2$: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

Substitute the values for an equi-convex lens: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$ $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{R} \right)$ $\frac{1}{f} = (\mu - 1) \left( \frac{2}{R} \right)$ So, the focal length is $f = \frac{R}{2(\mu - 1)}$.

The problem states that the focal length is greater than the radius of curvature of any of the surfaces, which is $R$. So, we have the condition $f > R$. $\frac{R}{2(\mu - 1)} > R$

Assuming $R > 0$ (radius of curvature is a magnitude, and for a real lens), we can divide both sides by $R$: $\frac{1}{2(\mu - 1)} > 1$

For a converging lens in air (which an equi-convex lens typically is, implying $f > 0$), the refractive index $\mu$ of the lens material must be greater than the refractive index of the surrounding medium (air, $\mu_{air} \approx 1$). Thus, $\mu > 1$, which means $\mu - 1 > 0$. Therefore, $2(\mu - 1)$ is positive. We can multiply both sides of the inequality by $2(\mu - 1)$ without reversing the inequality sign: $1 > 2(\mu - 1)$ $1 > 2\mu - 2$ Add 2 to both sides: $1 + 2 > 2\mu$ $3 > 2\mu$ Divide by 2: $\mu < \frac{3}{2}$ $\mu < 1.5$

Combining this result with the physical requirement that the refractive index of the lens material must be greater than that of the surrounding medium (air, $\mu \approx 1$) for it to be a converging lens, we have $\mu > 1$. So, the range for the refractive index is $1 < \mu < 1.5$.