Animated Solution for Physics - Optics: A ray of light entering from air into a denser medium of refractive index $\frac{4}{3}$, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle $\theta$ should be equal to

Snell's Law and Total Internal Reflection

Imagine a light ray embarking on a journey from the air into a denser medium, like a glass block. As it crosses the boundary, it bends, obeying the fundamental laws of optics. But the journey doesn't end there. It travels through the medium and strikes an adjacent surface, where it experiences a fascinating phenomenon: Total Internal Reflection (TIR). Our mission is to find the maximum angle at which this ray can enter the block and still undergo TIR at the second surface.

Applying Snell's Law

Let's start at the very beginning—the point where the light ray enters the denser medium. We apply Snell's Law, which relates the angle of incidence to the angle of refraction based on the refractive indices of the two media.

Snell's Law states:

In our specific scenario, the angle of incidence is $\theta$, and the angle of refraction is ${\theta }^{\prime }$. The refractive index of the denser medium is given as $\mu =\frac{4}{3}$. Substituting these values, we get our master equation for the first interface:

Unlocking Internal Angles

Now, let's look closely at the geometry inside the rectangular block. The normal to the first surface is horizontal, while the normal to the adjacent second surface is vertical. These two normals intersect at a perfect $90^{\circ }$ angle.

Together with the refracted ray, they form a right-angled triangle. In any right-angled triangle, the two acute angles must sum up to $90^{\circ }$. Therefore, the angle of refraction at the first surface (${\theta }^{\prime }$) and the angle of incidence at the second surface (${\theta }^{\prime \prime }$) are complementary:

This allows us to express the second angle of incidence purely in terms of the first angle of refraction:

The TIR Condition

For the light ray to be completely trapped inside the medium and undergo Total Internal Reflection at the second surface, the angle of incidence ${\theta }^{\prime \prime }$ must be greater than or equal to the critical angle. In the limiting case, we set it exactly equal to the critical angle, where the sine of the angle equals the reciprocal of the refractive index:

Substituting our geometric relation for ${\theta }^{\prime \prime }$, we get:

Using the fundamental trigonometric identity $\sin (90^{\circ }-x)=\cos x$, this simplifies beautifully to:

Finding the Maximum Angle

We now know the cosine of ${\theta }^{\prime }$, but our original Snell's Law equation requires the sine of ${\theta }^{\prime }$. We can easily bridge this gap using the Pythagorean identity ${\sin }^{2}x+{\cos }^{2}x=1$:

Finally, we bring this value back to our very first equation to solve for the incident angle $\theta$:

Notice how elegantly the $4$s cancel out, leaving us with:

Therefore, the maximum value of the incident angle $\theta$ is:

This perfectly matches option (a). By carefully tracing the ray's path and linking the geometry of the block with the laws of refraction, we've successfully unraveled the mystery of the light's journey.