In reading a scale reflected in a mirror by a telescope

In reading a scale reflected in a mirror by a telescope, the deflection $x$ is measured at a fixed scale distance $s$. How do small errors in the measured deflection $x$ affect the value of the angle $\alpha$?

Step 1: The geometric relation

When the mirror turns through an angle $\alpha$, the reflected ray turns through $2\alpha$ — the standard doubling of the mirror method. The spot on a scale at distance $s$ is therefore displaced by

\[ x = s\tan 2\alpha, \qquad s = \text{const}. \tag{1} \]

Step 2: Differentiate

\[ dx = s\cdot\frac{2}{\cos^{2}2\alpha}\,d\alpha = \frac{2s}{\cos^{2}2\alpha}\,d\alpha. \tag{2} \]

Step 3: Solve for the angular error

\[ d\alpha = \frac{\cos^{2}2\alpha}{2s}\,dx, \qquad \bigl|\Delta\alpha\bigr| \approx \frac{\cos^{2}2\alpha}{2s}\,\bigl|\Delta x\bigr|. \tag{3} \]

Step 4: Reading the answer

Two factors control the error.

The distance $s$. The angular error is inversely proportional to $s$: doubling the distance to the scale halves the error from a given reading error. This is why such instruments are set up with a long optical path.

The angle itself. The factor $\cos^{2}2\alpha$ is largest, namely $1$, at $\alpha = 0$, and decreases as $\alpha$ grows. So the same reading error produces the largest angular error near zero deflection:

\[ \bigl|\Delta\alpha\bigr|\Bigl|_{\alpha=0} \approx \frac{\bigl|\Delta x\bigr|}{2s}, \]

and a smaller one at larger $\alpha$, where the scale is more sensitive because $\tan$ is steepening. The relative error $\bigl|\Delta\alpha\bigr|/\alpha$ is worse still for small angles, since it grows without bound as $\alpha\to0$.

Step 5: Verification

Solving (1) explicitly gives $\alpha = \tfrac12\arctan\dfrac{x}{s}$. Differentiating: \[\frac{d\alpha}{dx} = \frac12\cdot\frac{1}{1+\bigl(x/s\bigr)^{2}}\cdot\frac1s = \frac{s}{2\bigl(s^{2}+x^{2}\bigr)}\]

\[\text{And } \cos^{2}2\alpha = \frac{1}{1+\tan^{2}2\alpha} = \frac{1}{1+(x/s)^{2}} = \frac{s^{2}}{s^{2}+x^{2}}\]

\[\frac{\cos^{2}2\alpha}{2s} = \frac{s^{2}}{s^{2}+x^{2}}\cdot\frac{1}{2s} = \frac{s}{2\bigl(s^{2}+x^{2}\bigr)} \;\;\checkmark\]