Focal Power of a Plane Interface

We know that when converging beam of rays are more converging after passing through an instrument, then we say it is of positive focal power. If angle of convergence doesn't change it is of zero focal power. Similarly negative focal power diverges.

Now if we consider an plane interface between two media. Let converging light rays pass through it from denser to rarer. Then the rays become less converging. That is negative power. But according to formula of focal power of surface when light rays pass from medium of refractive index n1 to medium n2, the focal power is n2-n1/R, where R is radius of curvature of interface. Here R=infinty. Therefore power is zero. Why a different answer?

As shown in the figure, Power is non zero. But the formula suggests that power is zero. Why?

1 Answer

In this example there is only one interface for which the lens makers formula, ignoring any particular sign convention, is $$\frac{n_1}{u} + \frac{n_2}{v} = \frac{n_2-n_1}{R}$$ where the object, distance $$u$$ from the interface, is situated in a material of refractive index $$n_1$$ and the image, distance $$v$$ from the interface, is observed from a material of refractive index $$n_2$$ and the radius of the interface is $$R$$.
Now if it a flat surface, $$R=\infty$$, the right hand side of the equation is zero which gives $$\frac u v = (-)\frac{n_1}{n_2}$$ which you might recognise as the real $$(u)$$ and apparent $$(v)$$ depth formula which is discussed here?