Question 121: Ray Optics

Question

What is the lateral shift of a ray refracted through a parallel-sided glass slab of thickness ' $h$ ' in terms of the angle of incidence ' $i$ ' and angle of refraction ' $r$ ', if the glass slab is placed in air medium?

Accepted Answer

We can derive the lateral shift (d) by considering the geometry of a ray entering a parallel-sided slab. HereÃ¢ÂÂs a stepÃ¢ÂÂbyÃ¢ÂÂstep explanation:
The slab has a thickness $$h$$ (measured perpendicular to the surfaces). When the ray enters the glass from air at angle $$i$$ (with respect to the normal), it refracts inside and makes an angle $$r$$.
Inside the slab, the ray travels a distance
$$L = \frac{h}{\cos r}$$
because the vertical component of the path must cover a distance $$h$$.
The ray inside the slab is shifted laterally (parallel to the surface). Its lateral displacement over the distance $$L$$ is
$$L \sin r = \frac{h \sin r}{\cos r}.$$
However, the emergent ray is parallel to the incident ray but not collinear. By constructing the appropriate geometry (extending the emergent ray backwards until it meets the incident ray's extension at the top interface), you find that the lateral shift between the incident ray and the emergent ray is given by
$$d = \frac{h \sin(i - r)}{\cos r}.$$
This expression correctly accounts for both the refraction inside the slab and the geometry of the displacement.
Thus, the correct answer is:
Option D: $$\frac{h \sin (i-r)}{\cos r}.$$

Answer

$\mathrm{h}$

Answer

$\frac{h \cos (i-r)}{\sin r}$

Answer

$\frac{\mathrm{h} 	an (\mathrm{i}-\mathrm{r})}{	an \mathrm{r}}$

Answer

$\frac{h \sin (i-r)}{\cos r}$

Ray Optics

What is the lateral shift of a ray refracted through a parallel-sided glass slab of thickness ' h ' in terms of the angle of incidence ' i ' and angle of refraction ' r ', if the glass slab is placed in air medium?

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