Step 1: Understanding the Concept:
The focal length of a lens depends on the refractive index of its material and the refractive index of the surrounding medium.
When a lens is immersed in a medium with a higher refractive index than its own material, its nature changes (a concave/diverging lens becomes convex/converging, and vice versa).
Step 2: Key Formula or Approach:
The Lens Maker's Formula is:
\[ \frac{1}{f} = \left( \frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For an equiconcave lens, by sign convention, the first surface is concave ($R_1 = -R$) and the second is convex towards the light ($R_2 = +R$).
Step 3: Detailed Explanation:
Given values:
Refractive index of lens, $\mu_g = 1.5$
Refractive index of medium (liquid), $\mu_l = 1.75$
Radii of curvature, $R_1 = -R$ and $R_2 = +R$
Using the Lens Maker's Formula:
\[ \frac{1}{f} = \left( \frac{1.5}{1.75} - 1 \right) \left( \frac{1}{-R} - \frac{1}{R} \right) \]
Calculate the relative refractive index term:
\[ \frac{1.5}{1.75} = \frac{150}{175} = \frac{6}{7} \]
So, $\left( \frac{6}{7} - 1 \right) = \left( \frac{6 - 7}{7} \right) = -\frac{1}{7}$
Calculate the radii term:
\[ \left( \frac{1}{-R} - \frac{1}{R} \right) = -\frac{2}{R} \]
Now, multiply these terms together to find $1/f$:
\[ \frac{1}{f} = \left( -\frac{1}{7} \right) \times \left( -\frac{2}{R} \right) = \frac{2}{7R} \]
Solving for focal length $f$:
\[ f = \frac{7R}{2} = +3.5 R \]
The positive sign of the focal length indicates that the lens behaves as a converging or convex lens.
Step 4: Final Answer:
It will act as a convex lens of focal length (3.5) R.