Problem 42 - Master Resource Book in JEE Main Physics

Question

In Young's double slit experiment, the seventh maximum with wavelength \(\lambda_{1}\) is at a distance \(d_{1}\) and the same maximum with wavelength \(\lambda_{2}\) is at a distance \(d_{2}\). Then, \(d_{1} / d_{2}=\) (a) \(\frac{\lambda_{1}}{\lambda_{2}}\) (b) \(\frac{\lambda_{2}}{\lambda_{1}}\) (c) \(\frac{\lambda_{1}^{2}}{R_{2}}\) (d) \(\frac{\lambda_{2}^{2}}{\lambda_{1}}\)

Step-by-Step Solution

Answer

(a) \(\frac{\lambda_1}{\lambda_2}\)

Step 1: Understand the Concept of Maxima in Double Slit Experiment

In Young's double slit experiment, maxima occur where the path difference is an integer multiple of the wavelength. The condition for the m-th maximum is given as \( d \sin(\theta) = m\lambda \), where \(d\) is the distance between the slits, \(\theta\) is the angle of the maxima, and \(m\) is an integer (order of the maxima).

Step 2: Relate Distance and Angle for Small Angles

For small angles, \( \sin(\theta) \approx \tan(\theta) \approx \frac{x}{L} \), where \(x\) is the distance on the screen from the central maximum, and \(L\) is the distance from the slits to the screen. Thus, \( d \frac{x}{L} = m\lambda \), giving \( x = \frac{m \lambda L}{d} \).

Step 3: Apply Formula to the Problem at Hand

The problem involves the seventh maximum \( (m=7) \) for the two different wavelengths, which is at distances \(d_1\) and \(d_2\) respectively. Using \( x = \frac{m \lambda L}{d} \), we have:\[ d_1 = \frac{7 \lambda_1 L}{d} \]\[ d_2 = \frac{7 \lambda_2 L}{d} \]

Step 4: Calculate the Ratio of the Distances for the Same Order

Taking the ratio \( \frac{d_1}{d_2} \), we have:\[ \frac{d_1}{d_2} = \frac{\frac{7 \lambda_1 L}{d}}{\frac{7 \lambda_2 L}{d}} \]The factors \(7\), \(L\), and \(d\) cancel out, giving:\[ \frac{d_1}{d_2} = \frac{\lambda_1}{\lambda_2} \]

Step 5: Select the Correct Option

The calculation shows that the correct option is where \( \frac{d_1}{d_2} = \frac{\lambda_1}{\lambda_2} \), which corresponds to option (a).

Key Concepts

Optical Interference

Optical interference is a fundamental concept that describes how waves, such as light, interact with each other. When two or more light waves overlap, interference can occur leading to either increased intensity in some regions (constructive interference) or decreased intensity (destructive interference) in others.
This is beautifully demonstrated in Young's double slit experiment, a classic physics experiment that illustrates interference in light waves. When light from a coherent source passes through two closely spaced slits, it results in an interference pattern of alternating bright and dark fringes. These fringes form due to the overlapping of light waves emerging from the slits, interfering constructively or destructively based on their path differences.
- **Constructive Interference:** Occurs when the path difference between two waves is an integer multiple of the wavelength, leading to bright fringes or maxima.
- **Destructive Interference:** Occurs when the path difference is a half-integer multiple of the wavelength, leading to dark fringes or minima.
Understanding optical interference is critical in explaining many optical devices and phenomena, such as holograms and anti-reflective coatings, that rely on controlling the interference of light waves.

Wavelength and Path Difference

Wavelength and path difference are key elements in comprehending the patterns in Young's double slit experiment. Wavelength, denoted by the Greek letter lambda (\lambda), represents the distance between consecutive peaks of a wave.
The path difference refers to the difference in distance that two waves travel before converging at a point. This difference determines how the waves interfere with each other.
In Young's experiment, the path difference (\delta) is crucial for determining whether interference will be constructive or destructive. It is directly dependent on the wavelength. The formula expressing this is:
\[d \, \sin(\theta) = m \lambda\]
Here, \(d\) is the distance between slits, \(\theta\) is the angle of interference, and \(m\) is the order of the maxima or minima.
By knowing the wavelength and measuring the path difference, one can predict at what angles and positions bright and dark fringes will appear on a screen. This relationship between wavelength and path difference highlights the wave nature of light.

Maxima and Minima Calculations

Calculating the positions of maxima and minima in interference patterns is crucial for understanding Young's double slit experiment. Maxima, or bright fringes, appear at positions where the path difference between beams is an integer multiple of their wavelength (\lambda). The formula used to determine these positions is:
\[x = \frac{m \lambda L}{d}\]
where \(x\) is the position on the screen from the central maximum, \(L\) is the distance to the screen, \(d\) is the slit separation, and \(m\) is the order of the bright fringe.
For minima, or dark fringes, the path difference is a half-integer multiple of the wavelength. These patterns occur between the bright fringes and follow a similar calculation but altered by a half-wavelength increment.
In practical applications, this can be used for precise measurements. For instance, calculating the distance between maxima can help determine changes in wavelength when the properties of the medium through which light travels are altered. Hence, mastering these calculations provides a deeper insight into the wave behavior of light.