Problem 27 - Arihant AIEEE Physics

Question

Word 'Newton' is printed on a paper and is placed on a horizontal surface below a cubical glass. The minimum value of refractive index of a cubical glass for whichletters are not visible from any of vertical faces of the glass, is: (a) \(\sqrt{3}\) (b) \(0.5\) (c) 1 (d) \(\sqrt{2}\)

Step-by-Step Solution

Answer

The minimum refractive index of the glass is \(\sqrt{2}\). This ensures all light is internally reflected. Answer: (d) \(\sqrt{2}\).

Step 1: Understanding the Concept

In this problem, we need to find the minimum refractive index so that the letters beneath the glass are not visible from any vertical face. This involves total internal reflection, which occurs when the light inside the denser medium (the glass) can't exit into the lighter medium (air).

Step 2: Total Internal Reflection Condition

Total internal reflection occurs when the angle of incidence is greater than the critical angle. The critical angle θ_c is given by \[\theta_c = \sin^{-1}\left(\frac{1}{{n}}\right)\]where \(n\) is the refractive index of the glass.

Step 3: Geometrical Consideration

For no visibility from the side, the light must undergo total internal reflection at all angles from vertical faces. The worst case is a diagonal path across the face. Thus, use the relation for critical angle at 45° incidence (since it covers diagonal travel): \[ \sin(45^\circ) = \frac{1}{n} \]

Step 4: Calculate the Minimum Refractive Index

Substituting \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\), we equate and solve:\[ \frac{1}{n} = \frac{1}{\sqrt{2}} \Rightarrow n = \sqrt{2} \].

Step 5: Verify the Result

Check the possibility of total internal reflection. With \(n = \sqrt{2}\), all angles of incidence greater than 45° ensure the light is internally reflected, making the letters underneath the cube invisible from vertical faces.

Key Concepts

Total Internal Reflection

Total internal reflection is the phenomenon where light traveling within a denser medium is completely reflected back into the medium instead of refracting out into a less dense medium. This happens when the light tries to move from a denser medium, like glass, into a less dense medium, such as air.
This only occurs under specific conditions:

• The light must travel from a denser medium to a less dense one.
• The angle of incidence must be greater than a certain critical angle.

The result is that the light does not leave the denser medium at all, but instead reflects entirely within it. This principle is used in fiber optics and can explain why certain objects become invisible when placed behind or within materials with particular refractive indexes.

Critical Angle

The critical angle is the minimum angle of incidence at which total internal reflection occurs. When light travels from a denser medium to a less dense one, if the angle of incidence is greater than this critical angle, the light gets completely reflected back into the medium.
To find the critical angle (\(\theta_c\)) for a medium like glass, we use the formula:
\[\theta_c = \sin^{-1}\left(\frac{1}{n}\right)\]
where \(n\) is the refractive index of the denser medium. If the angle of incidence is less than the critical angle, light will simply refract out of the medium. But when it's greater, total internal reflection happens, creating effects like invisibility of objects placed behind the material.

Light Refraction

Light refraction is the bending of light as it passes from one medium into another. This change in direction occurs because light travels at different speeds in different materials.
The degree to which the light is bent is determined by the refractive index of the materials. When light enters a denser medium, it slows down and bends towards the normal line (an imaginary line perpendicular to the surface). Conversely, when it exits to a less dense medium, it speeds up and bends away from the normal.
Refraction is responsible for many everyday phenomena, such as:

• The apparent bending of a straw in a glass of water.
• The way lenses focus light.
• The shimmering effect on hot roads.

Understanding refraction is key to solving many physics problems involving light behavior.

Physics Problem Solving

Physics problem solving often involves using fundamental principles, mathematical formulas, and logical reasoning to find solutions. Let's break down how we can approach a problem like the one given in the exercise.Firstly, recognize the relevant physics concepts involved. In this case:

• Identification of total internal reflection and critical angle.
• Using geometric reasoning to determine how light behaves within the glass cube.

Secondly, apply the formulas correctly. Understand how to derive or manipulate these formulas to extract the answer. For instance:

• Start with understanding how to calculate the critical angle.
• Determine the needed refractive index using the scenario given.
• Solve for the answer by equating the critical angle condition with known values like \(45^\circ\).

Finally, verify your results to ensure they make logical sense within the context of the problem. This structured approach will enhance understanding and improve problem-solving skills in physics.