JEE Main 2025 Jan 23 Shift 1, Physics Q13: Refraction at Plane and Spherical surfaces

Inside the slab the ray travels along a length $\ell$ such that the perpendicular thickness is $h=\ell \cos r$, so $\ell =\frac{h}{\cos r}$.

The lateral shift is measured perpendicular to the original (incident) ray. Geometry of the triangle formed by the incident-ray extension, the refracted ray, and the exit point gives:

$d=\ell \sin (i-r)=\frac{h\sin (i-r)}{\cos r}$.