Problem 64 - Master Resource Book in JEE Main Physics

Question

A ray of light travelling in glass \(\left(\mu=\frac{3}{2}\right)\) is incident on a horizontal glass air surface at the critical angle \(\theta_{c}\) If thin layer of water \(\left(\mu=\frac{4}{3}\right)\) is now poured on the glass air surface, the angle at which the ray emerges into air at the water-air surface is (a) \(60^{\circ}\) (b) \(45^{\circ}\) (c) \(\underline{90^{\circ}}\) (d) \(180^{*}\)

Step-by-Step Solution

Answer

The ray emerges at a \( 90^{\circ} \) angle into the air.

Step 1: Understanding Critical Angle

The critical angle \( \theta_c \) is the angle of incidence in the denser medium (glass) for which the angle of refraction in the less dense medium (air) is \(90^{\circ}\). From Snell's law, \( n_1 \sin \theta_c = n_2 \sin 90^{\circ} \) where \( n_1 = \frac{3}{2} \) and \( n_2 = 1 \). This gives \( \sin \theta_c = \frac{1}{n_1} = \frac{2}{3} \).

Step 2: Adding Water Layer

When water is added, the light first refracts from the glass into the water. Use Snell’s Law again: \( n_{glass} \sin \theta_c = n_{water} \sin \theta_w \). Substitute \( n_{glass} = \frac{3}{2} \), \( \sin \theta_c = \frac{2}{3} \), and \( n_{water} = \frac{4}{3} \). This gives \( \sin \theta_w = 1 \).

Step 3: Light Path in Water

Since \( \sin \theta_w = 1 \), it follows that \( \theta_w = 90^{\circ} \). This means that the refracted ray in water travels along the water-air interface and then emerges into the air without deviating from its path.

Key Concepts

Critical Angle

The critical angle is a concept in refraction that occurs when light travels from a denser medium to a less dense medium, such as from glass to air. At this angle, the light ray is refracted along the boundary and does not pass into the less dense medium.
To find the critical angle, we rely on Snell's Law, which is expressed as:

• \( n_1 \sin \theta_c = n_2 \sin 90^{\circ} \)

Here, \( n_1 \) is the refractive index of the denser medium (in our case, glass), and \( n_2 \) is the refractive index of the less dense medium (air). When the angle of refraction is \(90^{\circ}\), the path of the refracted light lies along the boundary. Calculating the critical angle ensures we know the exact point at which total internal reflection occurs.

Refraction

Refraction is the bending of light as it passes through a medium with a different refractive index. When light travels from one medium to another, like from glass to air or water, its speed and direction change.
This behavior is described by Snell's Law:

• \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \),

where \( \theta_1 \) is the angle of incidence and \( \theta_2 \) is the angle of refraction. The refractive indices, \( n_1 \) and \( n_2 \), depend on the nature of the materials. If the first medium is denser than the second, the light bends away from the normal.
In the original exercise, when a layer of water is added, the path of light changes once more. Now the light has to pass from the glass into the water before emerging into the air, showcasing both the complexity and the predictability of refraction.

Optics in Physics

Optics is a branch of physics that deals with the behavior and properties of light. It encompasses the study of reflection, refraction, and the various phenomena associated with light's interactions with matter.
Key concepts in optics include:

• Reflection: The bouncing back of light when it hits a surface.
• Refraction: The bending of light as it moves between media with different refractive indices.
• Dispersion: The splitting of light into its constituent colors, often seen through a prism.

Understanding optics allows us to develop technologies like lenses, glasses, and other optical instruments. In the exercise you encountered, the laws governing refraction and the critical angle help explain how light behaves as it transfers from one medium to another. Optics not only helps us explain these phenomena but also enables countless applications ranging from vision correction to sophisticated scientific instruments.