Problem 10 - Master Resource Book in JEE Main Physics

Question

Consider a ray of light incident from air into a slap of glass (refractive index \(n\) ) of width \(d\), at an angle \(\theta\). The phase difference between the ray reflected by the top surface of the glass and the bottom surface is [NCERT Exemplar] (a) \(\frac{4 \pi n d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{-1 / 2}+\pi\) (b) \(\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{12}\) (c) \(\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{1 / 2}+\frac{\pi}{2}\) (d) \(\frac{4 \pi d}{\lambda}\left(1-\frac{1}{n^{2}} \sin ^{2} \theta\right)^{12}+2 \pi\)

Step-by-Step Solution

Answer

The correct option is (a).

Step 1: Understand the Problem

We need to find the phase difference between light rays reflected from the top and bottom surfaces of a glass slab, while taking into account the refraction and thickness of the slab.

Step 2: Recognize Refraction and Total Internal Reflection

The light ray incident from air into the glass is refracted due to the change in medium. Understand that the phase difference is influenced by the optical path length and additional phase change upon reflection.

Step 3: Use Snell's Law to Find Refraction Angle

Snell's Law states \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \). For air to glass, this becomes \( \sin \theta = n \sin r \), where \(r\) is the angle of refraction.

Step 4: Calculate Optical Path Length in Glass

The optical path length (OPL) inside the glass is \( 2nd\cos r \), where \(n\) is the refractive index and \(d\) is the thickness of the glass.

Step 5: Convert Optical Path Length to Phase Difference

The phase difference \( \Delta \phi \) due to OPL is \(\frac{2 \pi}{\lambda} \times 2nd \cos r = \frac{4\pi nd \cos r}{\lambda} \).

Step 6: Use Snell's Law to Express Cosine in Terms of Known Variables

Using \( \sin^2 r = \frac{\sin^2 \theta}{n^2} \), we have \( \cos r = \sqrt{1 - \frac{\sin^2 \theta}{n^2}} \).

Step 7: Substitute and Simplify

Substitute \(\cos r\) into the phase difference formula: \[\Delta \phi = \frac{4\pi nd}{\lambda} \sqrt{1 - \frac{\sin^2 \theta}{n^2}}\].

Step 8: Additional Phase Change Due to Reflection

When light reflects off a boundary from a less to more dense medium, it undergoes an additional phase shift of \(\pi\). So, the total phase difference is: \[\Delta \phi_{total} = \frac{4\pi nd}{\lambda} \cdot \left(1 - \frac{\sin^2 \theta}{n^2}\right)^{-1/2} + \pi\].

Step 9: Identify Correct Option

Compare the expression for the total phase difference with the given options and identify the correct one, which is option (a).

Key Concepts

Snell's Law

When light travels from one medium to another, its speed changes, causing the light to bend. This bending, or refraction, is governed by Snell's Law. Snell's Law mathematically relates the angle of incidence and the angle of refraction like this:

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

Here, \(n_1\) and \(n_2\) are the refractive indices of the first and second mediums, respectively. \(\theta_1\) is the angle of incidence, while \(\theta_2\) is the angle of refraction. This law helps us understand how much the light ray will bend when it enters a new medium. If light enters glass from air, the equation becomes \(\sin \theta = n \sin r\), where \(r\) is the angle of refraction inside the glass. Hence, Snell's Law is crucial in determining how light behaves across different materials.

Optical path length

The optical path length (OPL) is a measure of how far light has traveled in terms of the phase it accumulates. It accounts not just for the geometric distance, but also for the refractive index of the medium. The OPL is calculated as:

• \(OPL = n \cdot d\),

where \(n\) is the refractive index and \(d\) is the physical thickness of the medium. When light travels through a medium like glass, the OPL is effectively increased due to the higher refractive index. For example, in a slab of glass, the OPL is given by \(2nd\cos r\), where \(\cos r\) is the cosine of the refraction angle. This equation shows that the OPL is dependent on the angle at which light passes through the medium. Understanding OPL is essential to determining the phase change of light as it enters and exits various materials.

Refraction and reflection

Light behaves astonishingly when it interacts with surfaces. When light hits a surface, two key phenomena occur: refraction and reflection.

• Refraction: This involves the bending of light as it passes from one medium to another. The light ray changes direction because the speed of light is different in various materials. Refraction is described by Snell's Law, and it determines how much the light ray bends.
• Reflection: Here, some of the light bounces back into the original medium. This occurs at the surface boundary. The extent of reflection depends on the angle of incidence and the refractive index difference between the two mediums.

Refraction and reflection together define how light is transmitted and redirected through different materials, crucial for creating lenses and optical instruments.

Refractive index

The refractive index of a material is a fundamental property that describes how light propagates through it. It is often denoted by the symbol \(n\), and it measures the extent to which the speed of light is reduced inside the medium, compared to in a vacuum. For glass, a typical refractive index might be around 1.5, meaning light travels 1.5 times slower than it does in a vacuum.

• A higher refractive index indicates slower light speed within the medium.
• The refractive index also influences the bending of light via Snell’s Law.

In the context of the given exercise, the refractive index is essential for calculating key parameters like the optical path length and understanding phase changes as light travels through the glass slab.

Phase change upon reflection

When light reflects off a boundary between two mediums with different refractive indices, a phase change can occur. If the light reflects off a denser medium (higher refractive index), an additional phase change of \(\pi\) (equivalent to half a wavelength) occurs.

• This phase shift is crucial in optical applications like interferometry and thin-film coatings.
• When combined with the geometric and refractive properties of the materials involved, it affects the total phase difference of the light wave.

In particular situations such as light reflecting from a glass slab, this phase change because of reflection must be added to any phase shifts that occur as a result of traversing the material itself. It's a key factor in determining the overall phase difference between different light paths, which is what we calculated in our exercise.