For fixed values of radii of curvature of lens, power of the lens will be _______.

Fill in the blank with the correct answer from the options given below

The correct option is P ∝(µ−1).

To understand why this relation holds, we can refer to the Lens Maker's Formula. This formula relates the focal length of a lens to the refractive index of its material and the radii of curvature of its two surfaces:

$$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

where:
• $f$ is the focal length of the lens,
• $\mu$ is the refractive index of the lens material relative to the surrounding medium,
• $R_1$ and $R_2$ are the radii of curvature of the two lens surfaces.

The power ($P$) of a lens is defined as the reciprocal of its focal length ($f$):

$$P=\frac{1}{f}$$

Substituting this definition of power into the Lens Maker's Formula, we get:

$$P=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

For fixed values of the radii of curvature ($R_1$ and $R_2$), the term $\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ becomes a constant.

Therefore, the power of the lens is directly proportional to $(\mu -1)$:

$$P\propto (\mu -1)$$