The power of a lens (biconvex) is 1.25 m$$^{-1}$$ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm and 40 cm respectively. The refractive index of surrounding medium:

We need to determine the refractive index of the surrounding medium ($$\mu_m$$) given the parameters of a biconvex lens operating inside that medium.

1. Identify the Given Parameters

• Power of the lens ($$P$$) = $$1.25\text{ m}^{-1}$$
• Refractive index of the lens material ($$\mu_l$$) = $$1.5$$
• Radius of curvature of the first surface ($$R_1$$) = $$20\text{ cm} = 0.2\text{ m}$$
• Radius of curvature of the second surface ($$R_2$$) = $$40\text{ cm} = 0.4\text{ m}$$

According to the standard Cartesian sign convention for a biconvex lens:

• The first refracting surface is convex, so its radius of curvature is positive: $$R_1 = +0.2\text{ m}$$.
• The second refracting surface is concave, so its radius of curvature is negative: $$R_2 = -0.4\text{ m}$$.

2. Apply the Lens Maker's Formula

The relationship between the power of a lens, its relative refractive index, and its radii of curvature is given by the Lens Maker's Formula:

$$P = \frac{1}{f} = \left(\frac{\mu_l}{\mu_m} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Substituting the known values into the formula:

$$1.25 = \left(\frac{1.5}{\mu_m} - 1\right) \left(\frac{1}{0.2} - \frac{1}{-0.4}\right)$$

3. Simplify and Solve for $$\mu_m$$

First, calculate the terms inside the geometric curvature bracket:

$$\left(\frac{1}{0.2} + \frac{1}{0.4}\right) = 5 + 2.5 = 7.5\text{ m}^{-1}$$

Substitute this value back into the main equation:

$$1.25 = \left(\frac{1.5}{\mu_m} - 1\right) \times 7.5$$

Isolate the bracket containing the refractive index variables:

$$\frac{1.25}{7.5} = \frac{1.5}{\mu_m} - 1$$

$$\frac{1}{6} = \frac{1.5}{\mu_m} - 1$$

Add 1 to both sides of the equation:

$$\frac{1.5}{\mu_m} = 1 + \frac{1}{6} = \frac{7}{6}$$

Now, rearrange to find $$\mu_m$$:

$$\mu_m = 1.5 \times \frac{6}{7} = \frac{3}{2} \times \frac{6}{7} = \frac{18}{14} = \frac{9}{7}$$

Conclusion

The refractive index of the surrounding medium is $$\frac{9}{7}$$. Based on the options provided in the problem layout, this directly matches Option B.

Correct Answer: Option B ($$\frac{9}{7}$$)