Ratio of Focal Length to Radius of Curvature for Convex Lens

To find the ratio of the focal length ($f$) to the radius of curvature ($R$) of the convex lens, we use the Lens Maker's Formula:

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

Given:

• Refractive index of the glass lens, $\mu = 1.4$
• The lens is a double convex (equiconvex) lens with equal radii of curvature on both sides.

Applying the standard Cartesian sign convention for a convex lens:

• $R_1 = +R$
• $R_2 = -R$

Substitute these values into the Lens Maker's Formula:

$\frac{1}{f} = (1.4 - 1)\left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right)$

$\frac{1}{f} = 0.4 \times \left( \frac{2}{R} \right)$

$\frac{1}{f} = \frac{0.8}{R}$

Rearranging to find the ratio of focal length to radius of curvature ($\frac{f}{R}$):

$\frac{f}{R} = \frac{1}{0.8} = \frac{10}{8} = 1.25$

Thus, the ratio of its focal length and radius of curvature is 1.25, which corresponds to Option D.