Problem 37 - Master Resource Book in JEE Main Physics

Question

If eye is kept at a depth \(h\) inside water of refractive index and viewed outaide, then the diameter of the cirele through which the outer objects become visible, will be (a) \(\frac{h}{\sqrt{\mu^{2}-1}}\) (b) \(\frac{h}{\sqrt{\mu_{2}+1}}\) (c) \(\frac{2 h}{\sqrt{\mu^{2}-1}}\) (d) \(\frac{h}{\sqrt{\mu^{2}}}\)

Step-by-Step Solution

Answer

The diameter is given by option (c) \( \frac{2h}{\sqrt{\mu^2 - 1}} \).

Step 1: Understanding the Problem

The problem involves finding the diameter of a circle through which objects outside the water are visible when viewed from inside water. This phenomenon is due to total internal reflection, and the circle is known as the circle of illumination.

Step 2: Applying Snell's Law

According to Snell's Law, the refractive index is used to find the critical angle for total internal reflection at the water-air interface. If \( \theta_c \) is the critical angle, then \( \sin(\theta_c) = \frac{1}{\mu} \) where \( \mu \) is the refractive index of water.

Step 3: Calculating the Critical Angle

The critical angle \( \theta_c \) can be calculated using \( \theta_c = \sin^{-1}\left(\frac{1}{\mu}\right) \). This angle is crucial for determining the boundary of the circle of illumination.

Step 4: Analyzing the Geometry

The geometry of the situation involves forming a cone of light within the water that refracts at the critical angle. The base of this cone, when extended up to the water's surface, creates the circle of illumination.

Step 5: Finding the Diameter of the Circle

The diameter of the circle at the surface, from geometry using the critical angle, can be given by \( 2h \tan(\theta_c) \). Substituting \( \tan(\theta_c) = \sqrt{\frac{\mu^2 - 1}{1}} \), we get the diameter as \( \frac{2h}{\sqrt{\mu^2 - 1}} \).

Step 6: Verifying the Answer

According to the calculations and considering the refraction effects, option (c) \( \frac{2h}{\sqrt{\mu^2 - 1}} \) is the correct choice.

Key Concepts

Snell's Law

To understand why the circle of illumination forms when looking out from underwater, we have to use a fundamental concept from optics: Snell's Law. Snell's Law describes how light behaves when passing between two mediums with different refractive indices, such as water and air.
The law states that:

• The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.

Mathematically, Snell's Law is expressed as:
\[\frac{\sin(\theta_i)}{\sin(\theta_r)} = \mu\]Where \( \theta_i \) is the angle of incidence, \( \theta_r \) is the angle of refraction, and \( \mu \) is the refractive index of the second medium with respect to the first.
In our scenario, a light ray traveling from water to air will bend away from the normal because water has a higher refractive index than air. When looking at the circle of illumination, Snell's Law helps us understand why light rays from outside form a boundary that makes a distinct circle when viewed underwater.

Critical Angle

The critical angle is a unique angle of incidence at which light, passing from a denser medium to a less dense medium, refracts along the boundary. Beyond this angle, light cannot pass through the boundary and instead reflects back into the denser medium.
To find the critical angle, we need to consider when the angle of refraction becomes 90 degrees. Using Snell's Law, the formula for the critical angle \( \theta_c \) is:
\[\sin(\theta_c) = \frac{1}{\mu}\]Where \( \mu \) is the refractive index of the denser medium (water, in this case).
Calculating the critical angle involves:

• Taking the inverse sine of the reciprocal of the refractive index.

This angle is fundamental when looking at optically dense materials like water, as it helps define the boundary of the circle of illumination through total internal reflection.

Total Internal Reflection

Total internal reflection occurs when a light ray within a denser medium hits the boundary with a less dense medium at an angle greater than the critical angle. At this point, instead of refracting out, the light reflects entirely back inside.
This phenomenon relies heavily on the concept of the critical angle. Once a light ray hits above this angle, no light will escape, and this principle is what creates the circle of illumination.

• The circle forms because light rays at angles smaller than the critical angle refract out, enlightening the boundary or 'circle' visible from inside.
• Rays greater than the critical angle are reflected back, and no light passes through, defining the edge of this circle.

In essence, total internal reflection is what enables the circle of illumination to form. It demonstrates how critical angles and Snell's Law work together to create unique optical phenomena observable in mediums like water.