Problem 29 - Master Resource Book in JEE Main Physics

Question

Two slits separated by a distance of \(1 \mathrm{~mm}\) are illuminated with red light of wavelength \(6.5 \times 10^{-7} \mathrm{~m}\). The interference fringes are observed on a screen place \(1 \mathrm{~m}\) from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (a) \(0.65 \mathrm{~mm}\) (b) \(1.63 \mathrm{~mm}\) (c) \(3.25 \mathrm{~mm}\) (d) \(4.88 \mathrm{~mm}\)

Step-by-Step Solution

Answer

The distance is \(1.63 \mathrm{~mm}\).

Step 1: Understanding the Fringe Formulas

To find the positions of dark and bright fringes on the screen, we use the formulas for constructive (bright) and destructive (dark) interference. For bright fringes (constructive interference), the position is given by \( y = \frac{m \lambda D}{d} \) where \( m \) is the order of the fringe, \( \lambda \) is the wavelength, \( D \) is the distance to the screen, and \( d \) is the slit separation. For dark fringes (destructive interference), the formula is \( y = \frac{(m + 0.5) \lambda D}{d} \).

Step 2: Calculate the Position of the Third Dark Fringe

Substitute \( m = 2 \) into the dark fringe formula since the first dark fringe corresponds to \( m = 0 \): \( y_3 = \frac{(2 + 0.5) \cdot 6.5 \times 10^{-7} \times 1}{1 \times 10^{-3}} = \frac{2.5 \times 6.5 \times 10^{-7}}{1 \times 10^{-3}} = 1.625 \times 10^{-3} \) meters.

Step 3: Calculate the Position of the Fifth Bright Fringe

Substitute \( m = 5 \) into the bright fringe formula: \( y_5 = \frac{5 \cdot 6.5 \times 10^{-7} \times 1}{1 \times 10^{-3}} = \frac{5 \times 6.5 \times 10^{-7}}{1 \times 10^{-3}} = 3.25 \times 10^{-3} \) meters.

Step 4: Calculate the Distance Between the Fringes

The distance between the third dark fringe and the fifth bright fringe is the difference in their positions: \( \Delta y = y_5 - y_3 = 3.25 \times 10^{-3} - 1.625 \times 10^{-3} = 1.625 \times 10^{-3} \) meters, which is equivalent to 1.63 mm.

Key Concepts

Young's Double Slit Experiment

Imagine shining a light through two tiny slits. When this happens, light waves spread out from each slit and overlap with each other on a screen placed a bit further away. This overlapping causes a pattern of bright and dark spots, known as interference fringes, to appear on the screen. Young's Double Slit Experiment is famous because it revealed that light behaves like a wave, similar to ripples on water.

Why are interference fringes important? Because they show where light waves combine at their peaks (bright spots) and where they cancel out (dark spots). These happen due to two key phenomena: constructive and destructive interference. With this experiment, Thomas Young showed how light can create such patterns, offering critical evidence for the wave theory of light and greatly expanding our understanding of physics and optics.

In practical terms, this experiment is still relevant today. It helps us understand various optical technologies, including lasers and fiber optics. By manipulating how waves interact, modern technology controls light in ways that are crucial for communication and other applications.

Constructive Interference

Constructive interference is a phenomenon where two or more waves combine to form a wave with a greater amplitude. In the context of Young's Double Slit Experiment, bright fringes on the screen are results of constructive interference. This happens when the path difference between the two waves coming from the slits is an integer multiple of the light's wavelength, denoted by \( m \), such as \( m = 0, 1, 2, \text{etc.} \)

Mathematically, the position of these bright fringes can be found using the formula:

• \( y = \frac{m \lambda D}{d} \)

Here, \( \lambda \) is the wavelength of the light, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. Whenever the condition for constructive interference is met, the waves amplify each other, leading to a bright spot.

Constructive interference is not just a cool physics trick; it's a principle leveraged in many technologies. For example, it is used in noise-canceling headphones, where sound waves are combined to eliminate unwanted noise. It also plays a role in the operation of various imaging devices and even in the design of energy-efficient lighting.

Destructive Interference

While constructive interference leads to bright spots, destructive interference causes darkness. Destructive interference happens when two light waves from the slits meet in such a way that their peaks and troughs align oppositely, canceling each other out.

In Young's Double Slit Experiment, this results in dark fringes, occurring when the path difference is a half-integer multiple of the wavelength, like \( m + 0.5 \). The formula to find the position of dark fringes is:

• \( y = \frac{(m + 0.5) \lambda D}{d} \)

This means the waves are exactly half a wavelength out of phase.

Destructive interference is also more than just an academic concept. It's applied in engineering, especially in minimizing signal interference in wireless communication and creating optical filters. Understanding this phenomenon helps scientists and engineers design ways to control and manipulate signals in highly precise ways.