​When a ray of light passes through a transparent glass slab, it does not deviate rather it gets laterally shifted. The perpendicular distance through which the ray is shifted is known as lateral shift.

Consider a transparent glass slab of thickness $t$. Let a ray of light travelling along path $OP$ is incident on the slab at an angle of incidence $i$. Then, the ray of light gets refracted in the denser medium (glass slab). The ray of light bends towards normal and travels along path $PQ$ such that the angle of refraction is $r$.

Since the two faces of the slab are parallel to each other, the ray of light $PQ$ strikes the surface at an angle of incidence equal to $r$. Finally, it emerges out of the slab along $QR$ making an angle of emergence equal to $i$.

Lateral Shift

Here, the emergent ray $QR$ is parallel to the incident ray $OP$ but the ray has been shifted through a perpendicular distance $QN$. This perpendicular distance $QN$ is known as lateral shift.

​In rt. angled $ΔPQN$, \[\angle QPN = i-r\] Thus, \[\sin(i-r)=\frac{QN}{PQ}\] \[∴QN=PQ\sin(i-r)\text{ __(1)}\]

In rt. angled $ΔPQM$, \[\cos r=\frac{PM}{PQ}\] \[PQ=\frac{PM}{\cos r}\] \[∴PQ=\frac{t}{\cos r}\text{ __(2)}\]

From $(1)$ and $(2)$, \[QN=\frac{t}{\cos r}\sin(i-r)\] \[d=\frac{t}{\cos r}\sin(i-r)\] This gives the lateral shift.

Now, if $i=90°$, \[d=\frac{t}{\cos r}\sin(90-r)\] \[d=\frac{t}{\cos r}\cos r\] \[d=t\] Thus, lateral shift will be maximum if $i=90°$.

[Also See: Dispersion]