Q.Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.29. A current in the coil produces a deflection of $3.5^\circ$ of the mirror. What is the displacement of the reflected spot of light on a screen placed $1.5\ \text{m}$ away?

A plane mirror rotated by $\theta$ deflects the reflected ray by $2\theta$ (the normal rotates with the mirror, doubling the shift on the reflection side).

1. Mirror deflection $\theta=3.5^\circ \Rightarrow$ reflected-ray deflection $=2\theta=7.0^\circ$. …

Why rotating the mirror doubles the ray's deflection

The law of reflection says the angle of incidence equals the angle of reflection, both measured from the mirror's normal. If the mirror rotates by $\theta$, its normal also rotates by $\theta$ - but the incident ray (from the fixed light source) doesn't move. So relative to the new normal, the angle of incidence has changed by $\theta$; by the law of reflection, the reflected ray moves by $\theta$ on the far side of the new normal too. Both shifts add up, so the reflected ray's total deflection is $2\theta$ - independent of the original angle of incidence.

Step 1: find the deflection of the reflected ray

Here light strikes the galvanometer mirror normally, and a current deflects the mirror by $\theta = 3.5^\circ$. The reflected ray therefore swings by

Step 2: find the displacement on the screen

The screen is at distance $L=1.5\ \text{m}$, perpendicular to the original (undeflected) ray direction. The reflected ray now makes an angle of $7.0^\circ$ with that original direction, so the spot's displacement $d$ is the opposite side of a right triangle with adjacent side $L$:

Using $\tan7.0^\circ \approx 0.1228$:

$d \approx 1.5\times0.1228 \approx 0.1842\ \text{m} \approx 18.4\ \text{cm}.$ …

Method: Relating Mirror Rotation to Reflected-Ray Deflection

This method solves any "optical lever" problem — a galvanometer mirror, a rotating mirror on a shaft — where a small mirror rotation must be converted into a spot displacement on a distant screen.

Steps

Step 1: Recall (and justify) the doubling rule

When a plane mirror rotates by a small angle $\theta$, its normal rotates by the same $\theta$. Since the incident ray from the fixed source doesn't move, the angle of incidence measured from the new normal has changed by $\theta$, and by the law of reflection the reflected ray shifts by another $\theta$ on the far side of the new normal. The two shifts add:

Step 2: Identify the given mirror rotation and double it

Read off the mirror's own rotation angle from the problem and multiply by 2 to get the actual angular swing of the reflected ray — this is the single most common place to slip (using $\theta$ instead of $2\theta$ in the next step).

Step 3: Convert the ray's angular swing into a linear displacement …