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

Step 1. A ray enters a glass slab of thickness $t$ at incidence angle $i$, refracting to angle $r$ as it travels through the slab (path $BC$), then leaves parallel to the original incident direction but displaced sideways by the lateral displacement $L$.

Step 2. Drop a perpendicular $CE$ from the exit point $C$ onto the (extended) undeviated original path; the angle between the incident-direction line and the actual exit path is $(i-r)$ (the net, uncancelled deviation, since the two individual refractions bend the ray towards, then away from, the normal by equal amounts $r$ and back by $i-r$ net overall).

Step 3. In the right triangle $\triangle BCE$: $\sin(i-r)=\dfrac{L}{BC}$, so $L=BC\sin(i-r)$. …

Drop a perpendicular from the exit point to the undeviated incident path and use two right triangles (one giving L, one …

• Confusing the net deviation angle (i-r) with the internal refraction angle r alone. …