A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is 30 cm and the refractive index of the lens material is 1.5, then the focal length of the lens (in cm) is

We begin with the Lens-maker’s formula for a thin lens placed in air:

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

Here

$$f=$$ focal length of the lens in air, $$\mu=$$ refractive index of the lens material with respect to air, $$R_1,\,R_2=$$ radii of curvature of the two spherical surfaces, measured from the lens toward the respective centres of curvature.

The lens given is plano-convex, so one surface is plane. For a plane surface the radius of curvature is infinite, that is

$$R_2=\infty \;\;\Longrightarrow\;\; \frac{1}{R_2}=0.$$

Substituting this into the Lens-maker’s expression we obtain

$$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-0\right)=(\mu-1)\frac{1}{R_1}.$$

The numerical values supplied are

$$\mu = 1.5, \qquad R_1 = 30\ \text{cm}.$$

Hence

$$\frac{1}{f} = (1.5-1)\,\frac{1}{30} = 0.5 \times \frac{1}{30} = \frac{1}{60}.$$

Taking the reciprocal gives

$$f = 60\ \text{cm}.$$

So the focal length of the plano-convex lens is $$60\ \text{cm}.$$

Hence, the correct answer is Option C.