Problem 83 - Master Resource Book in JEE Main Physics

Question

The Fraunhofer diffraction pattern of a single slit is formed in the focal plane of a lens of focal length \(1 \mathrm{~m}\). The width of slit is \(0.3 \mathrm{~mm}\). If third minimum is formed at a distance of \(5 \mathrm{~mm}\) from central maximum, then wavelength of light will be (a) \(5000 \mathrm{~A}\) (b) \(2500 \mathrm{~A}\) (c) \(7500 \AA\) (d) \(8500 \mathrm{~A}\)

Step-by-Step Solution

Answer

The wavelength of light is option (a) \(5000 \mathrm{~A}\).

Step 1: Understanding the Diffraction Condition

For a single slit diffraction pattern, the position of the minima is given by the condition \( a \sin \theta = m \lambda \), where \(a\) is the slit width, \(\theta\) is the angle of diffraction, \(m\) is the order of the minimum (1, 2, 3,...), and \(\lambda\) is the wavelength of light.

Step 2: Calculating Angle and Distance Relationship

Given that the third minimum \((m=3)\) is formed at a distance of \(5\, \text{mm}\) from the central maximum, we use small angle approximation: \( \tan \theta \approx \sin \theta \approx \theta \approx \frac{x}{f}\), where \(x = 5\, \text{mm} = 0.005\, \text{m}\) and \(f = 1\, \text{m}\). Hence, \(\theta = \frac{0.005}{1} = 0.005\).

Step 3: Substitute into Minima Condition

Using the diffraction condition, substitute the known values: \(a = 0.3\, \text{mm} = 0.0003\, \text{m}\), \(m=3\), and \( \theta \approx 0.005\). The minima condition becomes: \[ 0.0003 \times 0.005 = 3 \lambda \].

Step 4: Calculating Wavelength

Rearrange the equation to solve for \(\lambda\): \[ \lambda = \frac{0.0003 \times 0.005}{3} = \frac{0.0000015}{3} = 0.0000005 \, \text{m}\]. Convert this to Angstroms: \[ 1\, \text{m} = 10^{10} \, \text{Angstroms} \] therefore, \(\lambda = 0.0000005 \times 10^{10} = 5000 \, \text{A}\).

Key Concepts

Single Slit Diffraction

When light encounters an obstacle, like a single slit, it spreads out and creates a pattern of light and dark regions. This phenomenon is known as diffraction. In **single slit diffraction**, light passes through a narrow opening, or slit, and the light waves spread out. Instead of a simple shadow, an interference pattern is formed on a screen. This pattern has:

• Bright central maximum,
• Series of alternate dark and bright fringes.

The dark regions correspond to the minima (where light cancels out), and the brightest regions are maxima. This diffraction pattern arises due to the wave nature of light, and is influenced by several factors like the width of the slit, the wavelength of light, and the distance from the slit to the screen.

Wavelength Calculation

By analyzing the diffraction pattern from a single slit, we can calculate the wavelength of light using the formula for minima, which is: \[a\sin \theta = m\lambda\],where:

• \(a\) is the width of the slit,
• \(m\) is the order of the minimum (like first, second, third, etc.), and
• \(\lambda\) is the wavelength of light.

In the provided problem, we know:- **Slit width**, \(a = 0.3 \text{ mm} = 0.0003 \text{ m}\),- **Third minimum**, \(m=3\),- **Angle of diffraction**, \(\theta\) approximated using the small angle approximation.Using these, we rearrange the formula to solve for \(\lambda\):\[\lambda = \frac{a\theta}{m}\].Plugging in the known values, we find that the wavelength is \(5000 \text{ A}\). This conversion from meters to Angstroms recognizes that \(1 \text{ m} = 10^{10} \text{ A}\).

Diffraction Pattern Analysis

Analyzing a diffraction pattern helps us understand how light behaves when it encounters a single slit. Key Points:

• The central bright fringe (central maximum) is the most prominent in a single slit pattern.
• Each minimum corresponds to an angle where the path difference of light waves results in destructive interference, leading to the dark bands.
• This pattern is predicted by the Fraunhofer diffraction model, relevant when the light source and observation screen are far from each other.

The distances between maxima and minima can be measured, and the diffraction condition can then be used for calculations in experimental setups.

Angle of Diffraction

The **angle of diffraction** is crucial when studying diffraction patterns. It defines the direction of maximum or minimum intensity relative to the original direction of the wave.Small Angle Approximation:

• For small angles (when the distance between the slit and screen is much larger than the slit width), the angle \(\theta\) can be approximated using:\[\theta \approx \frac{x}{f}\]where \(x\) is the distance from the central maximum to the point of interest on the screen, and \(f\) is the focal length of the lens.
• In our exercise, it allows us to calculate the angle simply as \(\theta = 0.005\). This simplification makes it easier to solve the wavelength mysteries of light without complex trigonometry.

Understanding and applying these concepts helps in designing experiments and interpreting patterns in both educational and practical applications.