Q.A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle $i$ such that it is refracted out grazing the face AD. Find the value of angle $i$.

The key idea is the application of Snell's Law at two interfaces: first, to determine the critical angle for total internal reflection at face AD, and then to find the angle of incidence at face AB.

1. For the ray to graze face AD, the angle of incidence inside the glass at face AD must be the critical angle $C$. Applying Snell's Law at the glass-liquid interface (face AD):

   $n_{\text{glass}} \sin C = n_{\text{liquid}} \sin 90^\circ$

   $1.5 \sin C = 1.25 \times 1$

   $\sin C = \frac{1.25}{1.5} = \frac{5}{6}$

2. From the geometry of the rectangular slab, if the angle of refraction at face AB is $r$, then the angle of incidence at face AD is $90^\circ - r$. Thus, $C = 90^\circ - r$, which implies $r = 90^\circ - C$.

   Therefore, $\sin r = \sin(90^\circ - C) = \cos C$.

   Using $\cos C = \sqrt{1 - \sin^2 C}$:

   $\cos C = \sqrt{1 - \left(\frac{5}{6}\right)^2} = \sqrt{1 - \frac{25}{36}} = \sqrt{\frac{11}{36}} = \frac{\sqrt{11}}{6}$.

   So, $\sin r = \frac{\sqrt{11}}{6}$. …

To find the angle of incidence $i$, we need to work backward from the final condition: the ray grazing face AD. This condition tells us the angle of incidence inside the glass at face AD. We can then use the geometry of the rectangular slab to relate this angle to the angle of refraction at face AB. Finally, applying Snell's Law at face AB will give us the angle $i$.

The geometry of the slab and the ray path are shown in the figure:

1. Identify Refractive Indices and the Critical Angle Condition at Face AD

   The refractive index of the glass slab is $n_g = 1.5$.

   The refractive index of the surrounding liquid is $n_l = 1.25$.

   The ray of light, after refracting into the glass, strikes face AD and grazes out into the liquid. This means the angle of refraction in the liquid at face AD is $90^\circ$. The angle of incidence inside the glass at face AD is therefore the critical angle, let's call it $C$.

   For a ray going from a denser medium ($n_1$) to a rarer medium ($n_2$) at the critical angle $C$, Snell's Law gives $n_1 \sin C = n_2 \sin 90^\circ$, which simplifies to $\sin C = \frac{n_2}{n_1}$.

   Applying this to face AD, where light goes from glass ($n_g$) to liquid ($n_l$):

   $n_g \sin C = n_l \sin 90^\circ$

   $1.5 \sin C = 1.25 \times 1$

   $\sin C = \frac{1.25}{1.5} = \frac{125}{150} = \frac{5}{6}$

   Now, we find $\cos C$, which will be useful later:

   $\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \left(\frac{5}{6}\right)^2} = \sqrt{1 - \frac{25}{36}} = \sqrt{\frac{36 - 25}{36}} = \sqrt{\frac{11}{36}} = \frac{\sqrt{11}}{6}$

2. Relate Angles Using Geometry of the Rectangular Slab

   As the figure shows, the ray enters at face AB and exits grazing the adjacent face AD. Let $r$ be the angle of refraction at face AB (inside the glass), measured with respect to the normal to face AB.

   The angle of incidence at face AD is $C$, measured with respect to the normal to face AD.

   Since ABCD is a rectangular slab, faces AB and AD are perpendicular. Consequently, the normal to face AB is perpendicular to the normal to face AD.

   Consider the path of the refracted ray inside the glass. In the right-angled triangle formed by the ray and the two normals (one at AB, one at AD), the angles are related by:

   $r + C = 90^\circ$

   This means $C = 90^\circ - r$.

   Therefore, $\sin C = \sin(90^\circ - r) = \cos r$. …