A plano-convex lens (focal length $$f_2$$, refractive index $$\mu_2$$, radius of curvature R) fits exactly into a plano-concave lens (focal length $$f_1$$, refractive index $$\mu_1$$, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be:

Let the refractive index of air be $$\mu_0 = 1$$. The two lenses are in contact and, therefore, the light coming from air meets three successive refracting surfaces:

1. A plane surface separating air ($$\mu_0$$) and the plano-convex lens ($$\mu_2$$). 2. A common spherical surface of radius $$R$$ separating the plano-convex glass ($$\mu_2$$) and the plano-concave glass ($$\mu_1$$). 3. A plane surface separating the plano-concave glass ($$\mu_1$$) and air ($$\mu_0$$).

We apply, step by step, the formula for refraction at a spherical surface

$$ \frac{\mu_2}{v}-\frac{\mu_1}{u}= \frac{\mu_2-\mu_1}{R}, $$

stating first that for a plane surface $$R=\infty$$, so the right-hand side becomes zero and rays continue undeviated.

First (plane) surface
Object is at infinity ($$u=\infty$$). Because $$R=\infty$$, we get

$$ \frac{\mu_2}{v_1}-\frac{1}{\infty}=0 \;\;\Longrightarrow\;\; \frac{\mu_2}{v_1}=0 \;\;\Longrightarrow\;\; v_1=\infty . $$

After the first surface the rays are still parallel inside the $$\mu_2$$ glass.

Second (curved) surface
For this surface the object distance is $$u_2=\infty$$ (parallel rays). Using the same formula with $$\mu_1$$ (after) and $$\mu_2$$ (before) we have

$$ \frac{\mu_1}{v_2}-\frac{\mu_2}{\infty}= \frac{\mu_1-\mu_2}{R}. $$

The second term on the left vanishes, giving

$$ \frac{\mu_1}{v_2}= \frac{\mu_1-\mu_2}{R}\;\;\Longrightarrow\;\; v_2=\frac{\mu_1 R}{\mu_1-\mu_2}. $$

This distance $$v_2$$ is measured inside the $$\mu_1$$ glass from the curved surface to the point where the rays would meet if the glass extended that far.

Third (plane) surface
Here $$R=\infty$$ again, so

$$ \frac{\mu_0}{v_3}-\frac{\mu_1}{u_3}=0 \quad\Longrightarrow\quad \frac{1}{v_3}= \frac{\mu_1}{u_3}, $$

where $$u_3$$ is the object distance for this surface. The object for the third surface is the image formed by the second surface, therefore $$u_3 = v_2$$. Substituting,

$$ v_3=\frac{u_3}{\mu_1}= \frac{v_2}{\mu_1}= \frac{1}{\mu_1}\left(\frac{\mu_1 R}{\mu_1-\mu_2}\right)= \frac{R}{\mu_1-\mu_2}= -\frac{R}{\mu_2-\mu_1}. $$

The negative sign only indicates that the final image (for parallel incident rays) is on the opposite side of the combination, i.e. on the side where emergent light travels. The magnitude of this distance is the focal length $$F$$ of the entire combination:

$$ F=\frac{R}{\mu_2-\mu_1}. $$

Thus the equivalent focal length depends solely on the curved interface and equals the expression in Option A.

Hence, the correct answer is Option A.