Numerical Problems Based on Total Internal Reflection and Critical Angle for Class 12 Physics

The ray travels from air (rarer) to glass (denser), so it bends towards the normal. The angle of deviation is $\delta = i – r$.

Given $i = 45^\circ$ and $\delta = 15^\circ$:

Using Snell’s Law to find the refractive index of glass ($\mu_g$):

The critical angle $C$ for the glass-air interface is given by:

The relationship between the critical angle $C$ and the refractive index $\mu$ of a denser medium (with respect to air/vacuum) is $\sin C = \frac{1}{\mu}$.

First, find the refractive index of the glass ($\mu_g$) using its critical angle in air:

When immersed in water ($\mu_w = 1.33 = 4/3$), the new critical angle $C’$ is:

Let the refractive index of the core be $\mu$ and the angle of refraction at the entry face be $r$. The angle of incidence at the vertical core-cladding interface will be $(90^\circ – r)$.

For Total Internal Reflection at the vertical interface, the angle must be greater than or equal to the critical angle $C$:

Applying Snell’s law at the entry face: $\sin i = \mu \sin r \implies \sin r = \frac{\sin i}{\mu}$

Thus, the maximum angle of incidence is $i_{max} = \sin^{-1}(\sqrt{\mu^2 – 1})$.

The outside view is confined within a cone defined by rays reaching the fish at the critical angle of the water-air interface. The semi-vertical angle of this cone is exactly the critical angle $C$.

(Ray Diagram Tip: Draw a light source underwater. Show rays hitting the surface at varying angles. The ray hitting at $48.6^\circ$ will graze the surface. Rays hitting at angles greater than $48.6^\circ$ will reflect back entirely.)

First, determine the refractive index $\mu$ of the medium using the critical angle $C = 30^\circ$:

Now, calculate the speed of light in the medium $v$, given the speed of light in vacuum $c = 3 \times 10^8\text{ m/s}$:

When a ray grazes the surface separating the liquid and air, the angle of incidence in the liquid is the critical angle $C$. From the diagram’s geometry, you can extract $\sin C = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$.

The refractive index is $\mu = \frac{1}{\sin C}$. We also know $v = \frac{c}{\mu}$:

Substitute the value of $\sin C$ derived from the triangle in the diagram and multiply by $3 \times 10^8\text{ m/s}$ to get the final velocity.

For the source to be completely hidden, all light rays striking the surface outside the boundary of the opaque disc must suffer Total Internal Reflection. This means the rays hitting the exact edge of the disc must do so at the critical angle $C$.

From the geometry of the triangle formed by the depth $h$, radius $r$, and the light ray:

We know $\sin C = \frac{1}{\mu} = \frac{1}{5/3} = \frac{3}{5}$. If $\sin C = \frac{3}{5}$, using a right-angled triangle, the adjacent side is $\sqrt{5^2 – 3^2} = 4$. Therefore, $\tan C = \frac{3}{4}$.

Given $r = 1\text{ cm}$:

The ray PQ enters face BA normally ($i = 0^\circ$), so it passes undeviated and strikes the second face (AC) at an angle of incidence of $45^\circ$ (based on the diagram’s $45-90-45$ geometry).

Let’s check the critical angle $C$ for the prism material ($\mu = 1.5$):

Since the angle of incidence at face AC ($45^\circ$) is greater than the critical angle ($41.8^\circ$), the ray suffers Total Internal Reflection at face AC.

The ray will reflect at $45^\circ$, making it strike the third face (BC) normally ($90^\circ$). It will then emerge undeviated straight out from face BC.

When a ray of light “just grazes” the surface of separation between two media, the angle of refraction $r$ is $90^\circ$. The angle of incidence corresponding to this is the critical angle $C$.

Applying Snell’s Law:

Given $\mu_1 = \sqrt{2}$, $\mu_2 = 1$, and $\sin 90^\circ = 1$:

The angle of incidence must be $45^\circ$.