Question
Show that the deviation angle $\delta$ for a ray striking a thin converging lens at a distance $d$ from the principal axis is given by $\delta=d / f .$ Therefore, a ray is bent through an angle $\delta$ that is proportional to $d$ and does not depend on the angle of the incident ray (as long as it is par-axial). [ Hint: Look at the figure and use the small-angle approximation $\sin \theta \approx \tan \theta \approx \theta$ (in radians).]
Step 1
We label the point where the ray strikes the lens as $P$, the point where the ray converges as $Q$, and the center of the lens as $C$. We also label the angle of incidence as $\beta$ and the angle of deviation as $\gamma$. Show more…
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Show that the deviation angle $\delta$ for a ray striking a thin converging lens at a distance $d$ from the principal axis is given by $\delta=d / f$. Therefore, a ray is bent through an angle $\delta$ that is proportional to $d$ and does not depend on the angle of the incident ray (as long as it is parax-ial). [Hint: Look at the figure and use the small-angle approximation $\sin \theta=\tan \theta=\theta$ (in radians)].
When light passes through a prism, the angle that the refracted ray makes relative to the incident ray is called the deviation angle $\delta,$ Fig. $32-64 .$ Show that this angle is a minimum when the ray passes through the prism symmetrically, perpendicular to the bisector of the apex angle $\phi,$ and show that the minimum deviation angle, $\delta_{\mathrm{m}},$ is related to the prism's index of refraction $n$ by $$ n=\frac{\sin \frac{1}{2}\left(\phi+\delta_{\mathrm{m}}\right)}{\sin \phi / 2} $$ $\left[\right.$ Hint: For $\theta$ in radians, $\left.(d / d \theta)\left(\sin ^{-1} \theta\right)=1 / \sqrt{1-\theta^{2}} .\right]$
Angle of deviation for a ray incident at small angle on a thin prism of angle $\theta^{\circ}$ of refractive index $\mathrm{n}$ is (in figures): (a) zero (b) $\mathrm{n} \theta$ (c) $(\mathrm{n}-1) \theta$ (d) $(n+1) \theta$
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