Problem 26 - Master Resource Book in JEE Main Physics

Question

In Young's two slit experiment the distance between the two coherent sources is \(2 \mathrm{~mm}\) and the screen is at a distance of \(1 \mathrm{~m}\). If the fringe width is found to be \(0.03 \mathrm{~cm}\), then the wavelength of the light used is (a) \(4000 \mathrm{~A}\) (b) \(5000 \mathrm{~A}\) (c) \(5890 \mathrm{~A}\) (d) \(6000 \dot{A}\)

Step-by-Step Solution

Answer

The wavelength of the light used is \( 6000 \dot{A} \) (option d).

Step 1: Understanding the Given Variables

We have the distance between the two slits, which is the slit separation denoted by \( d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \). The distance from the slits to the screen, \( D = 1 \text{ m} \), and the fringe width, \( \beta = 0.03 \text{ cm} = 0.03 \times 10^{-2} \text{ m} \). We need to find the wavelength of light \( \lambda \).

Step 2: Using the Fringe Width Formula

In Young's double-slit experiment, the fringe width \( \beta \) is given by the formula: \[ \beta = \frac{\lambda D}{d} \], where \( \lambda \) is the wavelength of light, \( D \) is the distance to the screen, and \( d \) is the distance between the slits.

Step 3: Substitute Given Values Into the Formula

Substitute \( \beta = 0.03 \times 10^{-2} \text{ m} \), \( D = 1 \text{ m} \), and \( d = 2 \times 10^{-3} \text{ m} \) into the formula: \[ 0.03 \times 10^{-2} = \frac{\lambda \times 1}{2 \times 10^{-3}} \].

Step 4: Solve for Wavelength \( \lambda \)

Rearrange the equation to solve for \( \lambda \): \[ \lambda = \beta \times \frac{d}{D} = 0.03 \times 10^{-2} \times \frac{2 \times 10^{-3}}{1} = 0.03 \times 2 \times 10^{-5} = 6 \times 10^{-7} \text{ m} \].

Step 5: Convert Wavelength from Meters to Angstroms

Since 1 Angstrom (\( \dot{A} \)) = \( 10^{-10} \) meters, convert \( \lambda \) from meters to Angstroms: \[ 6 \times 10^{-7} \text{ m} = 6000 \dot{A} \].

Key Concepts

Fringe Width

Fringe width is a key concept in Young's double-slit experiment. It refers to the distance between consecutive bright or dark fringes on a screen. These fringes are the result of the constructive and destructive interference of light waves. The fringe width provides insights into the behavior of light upon passing through the slits. In essence, fringe width is calculated using the formula:
\[ \beta = \frac{\lambda D}{d} \]
where \( \beta \) represents fringe width, \( \lambda \) is the wavelength of the light used, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. Understanding fringe width helps students grasp how different variables affect interference patterns. When fringe width is larger, the fringes appear more spread out on the screen. Conversely, a smaller fringe width results in fringes that are closer together. By manipulating these variables, one can change the fringe pattern significantly.

• Fringe width depends on the wavelength of light, the slit separation, and distance to the screen.
• Increasing the wavelength or screen distance increases fringe width.
• Decreasing the slit separation also increases fringe width.

Wavelength Calculation

Calculating the wavelength of light in Young's double-slit experiment is a straightforward process. The key is to apply the formula for fringe width which relates directly to the wavelength:
\[ \beta = \frac{\lambda D}{d} \]
By rearranging this formula, you can solve for the wavelength \( \lambda \):
\[ \lambda = \frac{\beta d}{D} \]
Once you substitute the known values for fringe width \( \beta \), slit separation \( d \), and distance to the screen \( D \), you can compute the wavelength. In practice, this helps in determining the characteristics of the light used in the experiment.
Understanding how to calculate the wavelength allows students to connect theory with practical observations. It shows the precise relationship between measurable quantities such as fringe width and intrinsic properties like wavelength. This step is essential for any in-depth study of wave optics.

Wave Optics

Wave optics, also known as physical optics, studies the behavior of light in terms of its wave nature. This branch of optics is crucial when exploring phenomena like interference, diffraction, and polarization. Young's double-slit experiment is a classical representation of wave optics and demonstrates the principle of superposition.
As light passes through two closely spaced slits, it diffracts and the waves overlap. Where the peaks and troughs of these waves align, they interfere constructively or destructively, forming a distinct pattern of fringes.

• Constructive interference occurs when wave peaks align, resulting in bright fringes.
• Destructive interference happens when wave peaks align with troughs, leading to dark fringes.

Understanding wave optics is essential for comprehending more complex optical phenomena and is foundational for applied topics in physics like laser optics and optical engineering.

Coherent Sources

In Young's double-slit experiment, coherent sources are paramount. Coherence means that the light waves originating from the slits maintain a constant phase relationship. This consistency is necessary for clear and stable interference patterns.
Light sources that are coherent emit waves with the same frequency and a constant phase difference. In practical terms, lasers are common coherent light sources utilized in many optical experiments.

• Coherent sources ensure stable and distinct interference fringes.
• Lack of coherence would lead to jumbled and indistinct patterns.
• Coherence is crucial for achieving meaningful and observable results in wave interference experiments.

By using coherent sources, students can better explore and understand the principles underlying wave-based phenomena and gain insights into the behavior of light under different conditions.