Problem 47 - Master Resource Book in JEE Main Physics

Question

In a Young's experiment, two coherent sources are placed \(0.90 \mathrm{~mm}\) aprt and the fringes are observed one metre away. If it produces the second dark fringe at a distance of \(1 \mathrm{~mm}\) from the central fringe, the wavelength of monochromatic light used will be (a) \(60 \times 10^{-4} \mathrm{~cm}\) (b) \(10 \times 10^{-4} \mathrm{~cm}\) (c) \(10 \times 10^{-5} \mathrm{~cm}\) (d) \(6 \times 10^{-5} \mathrm{~cm}\)

Step-by-Step Solution

Answer

The wavelength of the monochromatic light is \(3.6 \times 10^{-5} \text{ cm}\), so the correct option is not among those given.

Step 1: Understanding the Formula

In the Young's double-slit experiment, the condition for dark fringes is related to the path difference between the light coming from the two sources. The formula for the position of the dark fringes can be expressed as: \( y = \dfrac{(2m + 1)\lambda D}{2d} \), where \( y \) is the distance of the dark fringe from the central fringe, \( D \) is the distance to the screen, \( d \) is the distance between the slits, \( m \) is the order number of the dark fringe (starting from zero), and \( \lambda \) is the wavelength of the light.

Step 2: Applying Given Values

We are given that the order of the dark fringe, \( m \), is 2, \( y = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \), \( D = 1 \text{ m} \), and \( d = 0.90 \text{ mm} = 0.90 \times 10^{-3} \text{ m} \). We need to find \( \lambda \). Substitute these values into the formula.

Step 3: Setting Up the Equation

The formula for the position of the dark fringe becomes \( y = \dfrac{(2 \times 2 + 1)\lambda \times 1}{2 \times 0.90 \times 10^{-3}} \). Simplify to solve for \( \lambda \).

Step 4: Calculating Wavelength

After replacing the values, we have: \( 1 \times 10^{-3} = \dfrac{5\lambda}{1.8 \times 10^{-3}} \). Rearrange to solve for \( \lambda \): \( \lambda = \dfrac{1 \times 1.8 \times 10^{-3}}{5} \).

Step 5: Final Calculation

Now perform the algebraic simplification: \( \lambda = \dfrac{1.8 \times 10^{-3}}{5} = 0.36 \times 10^{-3} \text{ m} \) which when converted to centimeters becomes \( \lambda = 3.6 \times 10^{-5} \text{ cm} \).

Key Concepts

Coherent Sources

In Young's double-slit experiment, a fundamental concept is the use of coherent sources. But what does coherent mean in this context? Coherent sources refer to light sources that emit waves with a constant phase difference and the same frequency. Imagine two lasers that emit beams of light that perfectly match each other's light waves. This synchrony is crucial, as it allows the diffraction and interference patterns observed in the experiment. The waves from these sources overlap and interact predictably, leading to unique patterns like bright and dark fringes on a screen. Without coherence, the pattern would look like random noise due to the fluctuating phase differences. So, during the experiment when you observe these ordered patterns, it's all due to the coherence of the light waves.

Dark Fringes Position Formula

The position of dark fringes in Young's experiment is essential to predicting the interference pattern. To determine where these dark bands appear, we use a specific formula involving the path difference of the light waves. The formula used is:\[ y = \dfrac{(2m + 1)\lambda D}{2d} \] - \( y \) represents the distance from the central fringe (where a bright spot appears on the screen) to the dark fringe.- \( D \) is the distance from the slits to the screen where the pattern is viewed.- \( d \) is the distance between the two slits in the setup.- \( m \) is the order of the dark fringe, starting from zero.- \( \lambda \) represents the wavelength of the light used.By understanding and manipulating this formula, you can predict where these dark areas will form. It's like having a map to guide you through the otherwise complex interference patterns.

Wavelength Calculation

Calculating the wavelength of light involved in Young's experiment is all about leveraging the relationships between the distances, slits, and fringe order.Following through with the exercise, the question tasked us with finding \( \lambda \) using known values:- \( m = 2 \) (order of the dark fringe)- \( y = 1 \times 10^{-3} \text{ m} \) (distance from central fringe)- \( D = 1 \text{ m} \) (distance to the screen)- \( d = 0.90 \times 10^{-3} \text{ m} \) (distance between slits)First, plug these into the dark fringe formula:\[ 1 \times 10^{-3} = \dfrac{(2 \times 2 + 1)\lambda \times 1}{2 \times 0.90 \times 10^{-3}} \]Solve for \( \lambda \) to find:\[ \lambda = \dfrac{1 \times 1.8 \times 10^{-3}}{5} = 0.36 \times 10^{-3} \text{ m} \]Convert meters to centimeters for easier comprehension in the options presented:\( \lambda = 3.6 \times 10^{-5} \text{ cm} \)Understanding how to calculate the wavelength helps explain the nature of the light used in the experiment, revealing part of the light's hidden characteristics through its pattern.