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

Explanation

Consider a glass slab of thickness t and refractive index n is kept in air medium. The path of the light is ABCD and the refractions occur at two points B and C in the galss 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 the right agle triangle `Delta BCE`,
`sin(i-r)=(L)/(BC),BC=(L)/(sin(i-r))`
In the right angle triangle `Delta BCF`,
`cos(r)=(t)/(BC),BC=(t)/(cos(r))`
Equating equations (1) and (2)
`(L)/(sin(i-r))=(t)/(cos(r))`
After rearranging,
`L=t(sin(i-rr)/(cos(r)))`
Lateral displacement depends upon the thickness of the slab. Thickher the slab, greater will be the lateral displacement. Greater the angle of incident, larger will be the lateral displacement.