## Question

###### As shown in Fig. $36-7$, a light ray $I O$ is incident on a small plane mirror. The mirror reflects this ray back onto a straight ruler $S C$ which is $1 \mathrm{~m}$ away from and parallel to the undeflected mirror MM. When the mirror turns through an angle of $8.0^{\circ}$ and assumes the position $M^{\prime} M^{\prime}$, across what distance on the scale will the spot of light move? (This device, called an optical lever, is useful in measuring small deflections.) When the mirror turns throug

As shown in Fig. $36-7$, a light ray $I O$ is incident on a small plane mirror. The mirror reflects this ray back onto a straight ruler $S C$ which is $1 \mathrm{~m}$ away from and parallel to the undeflected mirror MM. When the mirror turns through an angle of $8.0^{\circ}$ and assumes the position $M^{\prime} M^{\prime}$, across what distance on the scale will the spot of light move? (This device, called an optical lever, is useful in measuring small deflections.) When the mirror turns through $8.0^{\circ}$ the normal to it also turns through $8.0^{\circ}$, and the incident ray makes an angle of $8.0^{\circ}$ with the normal $N O$ to the deflected mirror $M^{\prime} M^{\prime}$. Because the incident ray $I O$ and the reflected ray $O R$ make equal angles with the normal, angle $I O R$ is twice the angle through which the mirror has turned, or $16^{\circ}$. Then $$ \overline{I R}=\overline{I O} \tan 16^{\circ}=(1.0 \mathrm{~m})(0.287)=29 \mathrm{~cm} $$