Question

Half angular width of the central maxima in the fraunhaufer diffraction due to slit width \(\frac{1200}{\sqrt{2}} \mu \mathrm{m}\) is \(45^{\circ}\). Then wavelength of the light is: (1) \(600 \mu \mathrm{m}\) (2) \(1200 \mu \mathrm{m}\) (3) \(\frac{600}{\sqrt{2}} \mu \mathrm{m}\) (4) \(\frac{1200}{\sqrt{2}} \mu \mathrm{m}\)

Short answer

The wavelength of the light is 600 μm.

Step-by-step solution

Understand the concept

For the single slit diffraction pattern, the angular position of the first minimum is given by the formula: \[ a \times \tan \theta = n \times \frac{u}{f } \] where \(a\) is the slit width, \(\theta\) is the angle, \(m\) is the order of the minimum (m=1 for the first minimum), \(\lambda\) is the wavelength of light.

Given Data

It's provided that the half angular width (angle to the first minimum) of the central maximum is \( 45^{\circ} \). This means \( \theta = 45^{\circ} \). The slit width \( a = \frac{1200}{\sqrt{2}} \mu m \).

Apply the formula

Using the formula for the first minimum: \[ a \sin \theta = \lambda \] Substitute the given values into the equation.

Solve for wavelength

We have \( a = \frac{1200}{\sqrt{2}} \mu m \) and \( \sin 45^{\circ} = \frac{1}{√2} \):\[ \frac{1200}{\sqrt{2}} \sin 45^{\circ} = \lambda \]\[ \frac{1200}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \lambda \]\[ \lambda = \frac{1200}{2} = 600 \mu m \].

Key concepts

Single Slit Diffraction

In optics, single slit diffraction occurs when light passes through a narrow slit and spreads out. This phenomenon happens because light behaves as a wave. When these waves encounter the edges of the slit, they bend and interfere with each other. This interference creates a pattern of bright and dark regions on a screen placed behind the slit. The bright regions are called maxima, and the dark regions are called minima. Understanding single slit diffraction is important as it is a basic concept in wave optics and helps explain more complex optical phenomena. In a single slit experiment, the width of the central maximum is determined by the slit width and the wavelength of the light used.

Wavelength Calculation

Calculating the wavelength of light in a single slit diffraction experiment requires knowing the slit width and the angle to the first minimum. The formula used is: \[ a \sin \theta = \lambda \] where \( a \) is the slit width, \( \theta \) is the angle of the first diffraction minimum, and \( \lambda \) is the wavelength of the light. For the given problem, the slit width \( a = \frac{1200}{\sqrt{2}} \, \mu \text{m} \), and the angle \( \theta \) is given as \( 45^{\circ} \). Plugging in the known values: \[ \lambda = \frac{1200}{\sqrt{2}} \sin 45^{\circ} \] Since \( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \), we get: \[ \lambda = \frac{1200}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1200}{2} = 600 \, \mu \text{m} \] This calculation demonstrates how the wavelength is directly linked to the physical properties of the setup and the behavior of the light wave.

Optics

Optics is the branch of physics that studies light and its interactions. It covers a wide range of phenomena involving reflection, refraction, and diffraction. Understanding optics allows us to explain how lenses form images, how waves interfere and create patterns, and how various optical devices function.

Within optics, diffraction is a significant topic as it deals with the spreading of light waves when they encounter obstacles. This helps in designing and understanding instruments like microscopes, telescopes, and even everyday items like cameras.
By mastering the fundamental principles of diffraction, students gain insights into more complex optical systems and applications. For instance, knowing how to calculate the wavelength of light from single slit diffraction experiments helps in fields ranging from material science to astronomy.