A concave lens (refractive index $=1 \cdot 5$) has both surfaces of same radius of curvature $R$. If it is immersed in a liquid of refractive index $1 \cdot 75$ it will act as a

The behavior of a biconcave lens immersed in a liquid depends on the relative refractive indices. Given the lens material index $n_L = 1.5$ and the liquid medium index $n_M = 1.75$, the lens maker's formula yields the focal length.

For a biconcave lens with equal radii of curvature $R$, the radii are $R_1 = -R$ and $R_2 = +R$ by standard sign convention. Substituting into the formula:

$\frac{1}{f} = \left(\frac{n_L}{n_M} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{1.5}{1.75} - 1\right) \left(\frac{1}{-R} - \frac{1}{R}\right)$

The refractive index term simplifies as $\frac{1.5}{1.75} - 1 = -\frac{1}{7}$, and the curvature term is $\frac{1}{-R} - \frac{1}{R} = -\frac{2}{R}$. Multiplying these gives:

$\frac{1}{f} = \left(-\frac{1}{7}\right) \left(-\frac{2}{R}\right) = \frac{2}{7R}$

Thus, $f = \frac{7R}{2} = 3.5R$. The positive focal length indicates the lens is converging, acting as a convex lens.

Final Answer: $\boxed{\text{C}}$