Suppose there is an equi-convex lens made of glass which has a focal length ($f$) of 30cm. Then, can we not say that the radius of curvature, $R$ of the lens is twice the focal length, i.e. $R = 60cm$?
Why do we need to use the Lens Maker's Formula for the same, which in fact gives a different result :
$$
\frac{1}{f}= \frac{\mu_2 – \mu_1}{\mu_1}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
$$
For $f=30 cm$, $\mu_1 = 1$, $\mu_2 = 1.5$, $R_1 = R$ and $R_2 = -R$, we get :
$$\frac{1}{30}= 0.5\times\frac{2}{R}$$
Or, $R=30cm$
How can one explain this? Also, how can the focal length be equal to the radius of curvature?
Best Answer
Intuitively, the radius of curvature has to depend on the index of refraction of the glass. If the index were $1$, the lens would have no effect at all. If the index were very high, say $10$, it would not take much curvature to get a given f'ocal length. Clearly we cannot just say the radius of curvature is twice the focal length.
You have applied the lens maker's formula correctly to your problem. There is no problem with $R=f$, in fact that is always true for $\mu_2=1.5\mu_1$. We are assuming a thin lens in this formula, so the diameter of the lens must be small compared with $R$. If the focal length is $30$ cm and the diameter of the lens is $1$ cm the thickness is twice the height of a circular segment. Given $R=30, c=1$ we have $h=R-\sqrt{R^2-(\frac c4)^2}=30-\sqrt{900-\frac 14}\approx 0.004$ so the lens is about $1$ mm thick