Q.The radius of curvature of the curved surface of a plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will

For a plano-convex lens ($R=20\ \text{cm}$, $n=1.5$), the lens maker's formula $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ gives $f=+40\ \text{cm}$ whichever face meets the incoming light first — flipping the lens relabels $R_1 \leftrightarrow R_2$ but the signed combination is unchanged. A converging (real) focus is formed either way, so the lens' …

Setting up

A plano-convex lens has one flat surface ($R=\infty$) and one convex surface of radius $R=20\ \text{cm}$, made of glass with $n=1.5$. The lens maker's formula is

where $R_1$ is the radius of the surface light meets first, and $R_2$ is the radius of the surface it meets second, signed by the New Cartesian Convention (positive if the centre of curvature lies on the side the light is travelling towards).

The question is really asking: does swapping which face is "first" change whether the lens converges or diverges light?

Orientation 1 — convex face towards the incoming light

The convex surface bulges toward the incident light, so its centre of curvature lies on the far side: $R_1=+20\ \text{cm}$. The flat surface is second: $R_2=\infty$.

Orientation 2 — flat face towards the incoming light

Now the flat surface is first: $R_1=\infty$. The convex surface is second, and it bulges away from the incident light, so its centre of curvature lies on the near side (the side light is coming from): $R_2=-20\ \text{cm}$.

Why the two orientations agree …

Method: Determining Whether a Lens's Behaviour Depends on Which Side the Object Is On

This method solves conceptual questions asking whether a lens (e.g. plano-convex, biconvex) acts as the same type of lens (converging/diverging) no matter which face the light enters through, or whether its behaviour flips with orientation.

Steps

Step 1: Compute the focal length for one specific orientation using the lens maker's formula

Assign $R_1$ and $R_2$ with the sign convention for light entering through a chosen face first (a flat face has $R=\infty$).

Step 2: Recognise the reversibility principle for a thin lens

If light instead enters through the OTHER face first, the roles of $R_1$ and $R_2$ swap, and both surfaces' sign labels flip together (what was measured "towards the outgoing side" is now measured "towards the incident side", and vice versa for the other surface). Because both signs flip together, the combination $\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ comes out numerically identical — so $1/f$, and hence $f$ itself (magnitude AND sign), does not change with orientation for a genuine thin lens.

Step 3: Read off the lens type from the sign of $f$

A positive $f$ (as for glass, $n>1$, with a net convex effect) means a converging lens; this sign was shown in Step 2 to be orientation-independent, so the lens is converging for an object on EITHER side. …