Problem 38 - Master Resource Book in JEE Main Physics

Question

A small bulb is placed at the bottom of a tank containing water to a depth of \(80 \mathrm{~cm}\). What is the area of the surface of water through which light from the bulb ean emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.) [NCERT] (a) \(4.6 \mathrm{~m}^{2}\) (b) \(3.2 \mathrm{~m}\) (c) \(5.6 \mathrm{~m}^{2}\) (d) \(2.6 \mathrm{~m}^{2}\)

Step-by-Step Solution

Answer

The area of the surface of water through which light emerges is approximately \(2.6 \ m^2\), option (d).

Step 1: Understand the Concept of Critical Angle

When light travels from a denser medium (water) to a rarer medium (air), it bends away from the normal. The critical angle is reached when the angle of refraction is 90 degrees, and beyond this angle, total internal reflection occurs. We need the critical angle to find the area of the light emerging.

Step 2: Calculate the Critical Angle

The critical angle can be calculated using Snell's Law. The formula for the critical angle \( \theta_c \) is given by \( \sin(\theta_c) = \frac{1}{n} \), where \( n \) is the refractive index of water. So, \( \sin(\theta_c) = \frac{1}{1.33} \). Calculate \( \theta_c \).

Step 3: Determine the Critical Angle Value

Using the value from Step 2, we calculate \( \theta_c = \sin^{-1}\left(\frac{1}{1.33}\right) \). This comes out to approximately \( 48.75^\circ \).

Step 4: Relate Critical Angle to Light Emergence Area

Only light within the angle \( \theta_c \) will refract out of the water. The critical angle forms the apex of a cone of emerging light, with the bulb at its tip. The surface area through which the light emerges can be traced by projecting this cone onto the water surface.

Step 5: Calculate Area Through Geometry

The cone forms a circle on the surface. To find the radius \( r \), use \( r = d \cdot \tan(\theta_c) \), where \( d = 0.8 \) m is the depth of the water. Calculate \( r \).

Step 6: Evaluate the Radius

Substitute the values: \( r = 0.8 \cdot \tan(48.75^\circ) \). This gives \( r \approx 0.907 \) m.

Step 7: Calculate the Area of the Circle

The area of the circle \( A \) is given by \( \pi r^2 \). Compute \( A = \pi (0.907)^2 \).

Step 8: Finalize the Area Value

Calculate the area, which is approximately \( \pi \times 0.823 \approx 2.584 \), rounding off gives the area as approximately \( 2.6 \ m^2 \).

Step 9: Select the Correct Answer from Choices

Based on the calculations, the area of the surface through which the light emerges is closest to option (d) \(2.6 \ m^2\).

Key Concepts

Total Internal Reflection

Total Internal Reflection is an exciting phenomenon. It occurs when light travels from a denser medium, like water, to a less dense medium, like air. During this process, the light bends away from the normal line at the surface. If the angle exceeds a certain limit, known as the critical angle, the light reflects entirely back into the denser medium instead of refracting out. This is what we call Total Internal Reflection.
A few real-world examples include the shimmering reflections seen in swimming pools or fiber optic cables used for high-speed internet.

• Critical Angle: The minimum angle of incidence at which Total Internal Reflection occurs.
• Denser Medium: Water, glass, or any substance where light travels slower.
• Rarer Medium: Air, where light travels faster.

Understanding this phenomenon helps in visualizing how light behaves at different surface boundaries. It's not just a theoretical concept, but a practical phenomenon useful in technology and nature.

Refractive Index

The refractive index is a crucial concept in optics. It measures how much light 'bends' when entering a new medium. The refractive index, denoted by the symbol 'n', is calculated by dividing the speed of light in vacuum by the speed of light in the medium. So, the formula is given by: \[ n = \frac{c}{v} \] where:

• \(c\) is the speed of light in a vacuum.
• \(v\) is the speed of light in the medium, such as water or glass.

For example, water has a refractive index of 1.33, which means light travels slower in water compared to a vacuum. This number also defines the extent to which light bends inside water. It influences the critical angle, thus affecting Total Internal Reflection. Having a higher refractive index means more bending of light, which is significant when designing lenses and optical devices.

Snell's Law

Snell's Law is fundamental when studying light refraction and reflection. It defines the relationship between the angles and the refractive indices of the two media through which light passes. The classic formula for Snell's Law is: \[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \] Here,

• \(n_1\) and \(n_2\) are the refractive indices of the two different media.
• \(\theta_1\) is the angle of incidence.
• \(\theta_2\) is the angle of refraction.

Snell's Law is essential to finding the critical angle for Total Internal Reflection. By substituting \( \theta_2 \) with 90 degrees, we can calculate the critical angle. For water with a refractive index of 1.33, applying Snell's Law lets us determine the angle beyond which Total Internal Reflection occurs. It's an instrumental principle widely used in optics and physics to describe light behavior across boundaries and interfaces.