Name: Ray Diagram: Lenses
Uploaded: 2019-07-19
Duration: 6 min 35 s
Description: In this video, let us discuss how ray diagrams for lenses can be used to determine the size and location of a refracted image. Get in to know more!

Question

Obtain the equation of lateral displacement of light passing through a glass slab.

Accepted Answer

Consider a glass slab of thickness $t$ and refractive index $n$ s kept in air medium. The path of the light is ABCD and the refractions occur at two points $B$ and $C$ in the glass slab. The angles of incidence $i$ and refraction $r$ are measured with respect to the normal $N_{1}$ and $N_{2}$ at the two points B and C respectively. The lateral displacement L is the perpendicular distance $CE$ drawn between the path of light and the undeviated path of light at point C.

In right-angled triangle $BCE$

$\sin (i - r) = \frac{L}{BC} or BC = \frac{L}{\sin (i - r)}$ $\frac{L}{\sin (i - r)} = \frac{t}{\cos r} or L = t [\frac{\sin (i - r)}{\cos r}]$ ....................(1)

In the right angle triangle $BCF$

$\cos r = \frac{t}{BC} or BC = \frac{t}{\cos θ}$ .............(2)

Equating equations (1) and (2) and rearranging,, we get, $\frac{L}{\sin (i - r)} = \frac{t}{cor r} or L = t [\frac{\sin (i - r)}{\cos r}]$

Lateral displacement depends upon the thickness of the slab. Thicker the slab, greater will be the lateral displacement. Greater the angle of incident, larger will be the lateral displacement.

Obtain the equation of lateral displacement of light passing through a glass slab.

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