Problem 28 - Arihant AIEEE Physics

Question

In a tank filled with water of refractive index \(5 / 3\), a point source of light is placed \(4 \mathrm{~m}\) below the surface of water. To cut-off all light coming out of water from the source, what should be the minimum diameter of a disc, which should be placed over the source on the surface of water ? (a) \(1 \mathrm{~m}\) (b) \(4 \mathrm{~m}\) (c) \(3 \mathrm{~m}\) (d) \(6 \mathrm{~m}\)

Step-by-Step Solution

Answer

Option (d) \(6 \mathrm{~m}\) is the correct answer.

Step 1: Understand Total Internal Reflection

When light passes from a denser to a rarer medium, it may undergo total internal reflection if the angle of incidence exceeds a critical angle. In this exercise, the point light source is under water looking to escape into air, which are the denser and the rarer medium respectively.

Step 2: Calculate the Critical Angle

The critical angle \(\theta_c\) for water-air interface can be found using \(\sin \theta_c = \frac{n_2}{n_1}\), where \(n_1 = \frac{5}{3}\) (refractive index of water) and \(n_2 = 1\) (refractive index of air). Therefore, \[ \sin \theta_c = \frac{1}{5/3} = \frac{3}{5} \]. Thus, \[ \theta_c = \sin^{-1}\left(\frac{3}{5}\right). \]

Step 3: Determine the Circle of Illumination

The radius \(r\) of the circle of illumination on the water surface can be found using geometry. Considering the depth \(d = 4 \mathrm{~m}\), the relation is \( r = d \tan \theta_c \). Use \( \tan \theta_c = \frac{\sin \theta_c}{\sqrt{1-\sin^2 \theta_c}} \). Substitute \( \sin \theta_c = \frac{3}{5} \), \( \tan \theta_c = \frac{3/5}{\sqrt{1-(3/5)^2}} = \frac{3/5}{4/5} = \frac{3}{4} \), so \( r = 4 \times \frac{3}{4} = 3 \mathrm{~m} \).

Step 4: Calculate Minimum Diameter of the Disc

The diameter of the disc must be at least twice the radius of the circle of illumination to cover all the light. So the minimum diameter required is \(2r = 2 \times 3 = 6 \mathrm{~m}\).

Step 5: Compare With the Given Choices

Given options: (a) \(1 \mathrm{~m}\), (b) \(4 \mathrm{~m}\), (c) \(3 \mathrm{~m}\), (d) \(6 \mathrm{~m}\). The calculated minimum diameter is \(6 \mathrm{~m}\), which matches option (d).

Key Concepts

Critical Angle

The concept of Critical Angle is key to understanding Total Internal Reflection. When light travels from a denser medium to a rarer medium, it bends away from the normal. There exists a specific angle, known as the Critical Angle, at which light refracts along the boundary and not out of the denser medium. Beyond this angle, light will entirely reflect back into the denser medium, rather than refract into the rarer medium.

This angle depends on the refractive indices of the two media. It can be calculated using Snell's Law, where the sine of the Critical Angle (\(\theta_c\)) is expressed as the ratio of the refractive indices:

\[ \sin \theta_c = \frac{n_2}{n_1} \]

In our example, this critical refractive event determines the extent to which light from a source under water (denser medium) will stay underwater instead of escaping into the air (rarer medium).

Refractive Index

The Refractive Index is a fundamental concept that describes how light propagates through a material. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the material. This concept explains why light refraction occurs when transitioning between different media.

Mathematically, the refractive index (\(n\)) is:

\[ n = \frac{c}{v} \]

where \(c\) is the speed of light in a vacuum, and \(v\) is the speed of light in the medium.

• A higher refractive index indicates a greater slowing of light.
• Different materials have different indices based on their optical density.

For instance, water has a refractive index of \(\frac{5}{3}\), indicating that light travels slower in water than in air. This slower velocity results in light bending or refracting at the boundary/interface between water and air.

Circle of Illumination

The Circle of Illumination is a useful concept in optics, especially in situations involving water or clear boundaries. It's the area on the surface where light escapes from an underwater source and transitions to the air, defined when light emerges from the denser medium.

To calculate it, geometry aids us through the critical angle \(\theta_c\). For a point source beneath water, the reach of light is limited, forming a circle at the surface. The formula to find the radius \(r\) of this circle is:

\[ r = d \tan \theta_c \]

where \(d\) is the depth of the light source below the surface.

• Understanding this dimension helps us determine the light spread at the boundary.
• In our specific problem, this circle dictates the size of a physical block or disc needed to prevent light from escaping.

Geometry of Optics

The study of the Geometry of Optics combines mathematical principles with physical behavior of light. It covers how light rays interact with materials, such as reflecting, refracting, or being absorbed. In simple terms, it uses angles and distances to visualize and predict the path of light.

In our context, understanding geometric relationships helps determine how light behaves as it exits the water. By knowing the depth and angle, we calculate the spread of illumination on the surface. Key geometric principles used here include:

• Calculating with trigonometric functions like sine and tangent.
• The relationship between angle, depth, and surface distance in determining light pathways.

These principles enable practical applications, such as how much of a surface needs coverage to prevent light from passing through or how to position objects to control light patterns.