The spherometer is used to measure radius of curvature of spherical surfaces (e.g., spherical lens and mirror). Its working principle is same as that of a screw gauge.

A spherometer has three fixed legs and an adjustable central legs. The central leg is has a micrometer head on its top. The micrometer has a main (vertical) scale and a circular scale.

The pitch ($p$) of a spherometer is the distance moved in one complete revolution of the circular scale. The least count (LC) of a spherometer of pitch $p$ and having $N$ divisions on the circular scale is given by \begin{align} \text{Least count}=\frac{p}{N}. \end{align} Usually, p = 1 mm, N=100, and LC = 0.01 mm.

The tips of three fixed legs touches the spherical surface to be measured. These tips forms a plane. They lies on an equilateral triangle of side $l$. The distance ($l$) between tips of fixed legs is measured with the help of a scale.

Initially, the spherometer is placed on a flat surface. Its central legs is adjusted (moved) so that tip of the central leg lies at the centroid of the equilateral triangle formed by fixed legs. In this condition, the distance between the tip of central leg and any fixed leg is $r$. From geometry, \begin{align} r\cos30=\frac{l}{2} \end{align} which gives \begin{align} r=\frac{l}{\sqrt{3}} \end{align}

Now, spheromoeter is placed on the spherical surface of radius of curvature $R$. All fixed legs shall touch the surface. The central leg is adjusted so that its tip also touches the spherical surface. The distance $h$ between the plane containing tips of three fixed legs and the tip of central leg is measured with the help of main scale and the circular scale. The central leg can be moved up or down from the zero mark on the vertical scale.

From Pythagoras triangle \begin{align} R^2 &=r^2+(R-h)^2 \end{align} Simplify and substitute $r=l/\sqrt{3}$ to get \begin{align} R&=\frac{l^2}{6h}+\frac{h}{2} \end{align}

**Problem:** The pitch of a spherometer is 1 mm and there are 100 divisions on its disc. It reads 3 divisions on the circular scale above zero when it is placed on a plane glass plate. When it rests on a convex surface, it reads 2 mm and 63 divisions on a circular scale. If the distance between its outer legs is 4 cm, the radius $R$ of curvature of the convex surface is

- 4 cm
- 2.6 cm
- 10.4 cm
- 6.3 cm

**Solution:** The least count of the spherometer is
\begin{align}
\text{LC}&=\frac{p}{N}=\frac{1}{100}=0.01\;\text{mm}
\end{align}
The zero error of the spherometer is its measurement on the plane surface, which is 3 divisions on the circular scale. Thus, zero error is
\begin{align}
\text{Zero error}=3\times\text{LC}=0.03\;\text{mm}.
\end{align}
The measured value of $h_m$ is 2 divisions on the main scale and 63 divisions on the circular scale i.e.,
\begin{align}
h_m&=2+63\times\text{LC}=2.63\;\text{mm}
\end{align}
The zero correction gives
\begin{align}
h &=h_m-\text{Zero error}\\
&=2.63-0.03 \\
&=2.60\;\text{mm}=0.26\;\text{cm}
\end{align}
The distance between fixed legs is $l=4$ cm. Apply the spherometer formula to get the radius of curvature of the convex surface
\begin{align}
R &=\frac{l^2}{6h}+\frac{h}{2}\\
&=\frac{4^2}{6(0.26)}+\frac{0.26}{2} \\
&=10.4\;\text{cm}.
\end{align}

**Question:** How will you find the focal length of lens by using a spherometer?

**Answer:** The focal length of a lens is given by the Lens maker's formula
\begin{align}
\frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
\end{align}
The spherometer is used to measure the radius of curvatures $R_1$ and $R_2$ of two surfaces of the lens. The refractive index $\mu$ of the lens material is known.

**Question:** What do you mean by radius of curvature of a surface?

**Answer:** It is the radius of the sphare from which the surface is cut. You can cut a glass sphere of radius $R$ to get a plano-convex lens. The radius of curvature of the curved surface of the lens is $R$ and that of the plane surface is infinity.