Problem 21 - Master Resource Book in JEE Main Physics

Question

In a Young's experiment, one of the slits is covered with a transparent sheet of thickness \(3.6 \times 10^{-3} \mathrm{~cm}\) due to which position of central fringe shifts to a position originally occupied by 30 th fringe. The refractive index of the sheet, if \(\lambda=6000 \AA\), is (a) \(1.5\) (b) \(1.2\) (c) \(1.3\) (d) \(1.7\)

Step-by-Step Solution

Answer

The refractive index of the sheet is 1.2 (option b).

Step 1: Understand the Problem

In Young's experiment, one of the slits is covered with a transparent sheet, causing the central fringe to shift to the position of the 30th fringe. We need to determine the refractive index of the sheet given its thickness and wavelength of light used.

Step 2: Write Down the Key Relations

The change in the path difference due to the introduction of the sheet is equal to an additional phase difference of entire fringes.The phase change introduced by the sheet is given by the formula: \[ \Delta x = (\mu - 1)t \] where \( \mu \) is the refractive index, \( t \) is the thickness of the sheet, and \( \Delta x \) is the path difference change which causes the shift in fringes.

Step 3: Express Path Difference in Terms of Fringe Shift

Since the central fringe shifts to where the 30th fringe was, the path difference introduced must be equal to 30 wavelengths:\[ (\mu - 1) t = n \lambda \]Here, \( n \) is 30, \( \lambda \) is the wavelength, and \( t \) is the thickness converted into meters.

Step 4: Substitute Given Values

Convert the thickness into meters \(( t = 3.6 \times 10^{-3} \text{ cm} = 3.6 \times 10^{-5} \text{ m})\) and the wavelength into meters \(( \lambda = 6000 \AA = 6000 \times 10^{-10} \text{ m})\). Substitute these values into the equation:\[ (\mu - 1) \times 3.6 \times 10^{-5} = 30 \times 6000 \times 10^{-10} \]

Step 5: Solve for the Refractive Index

Simplify and rearrange the equation to solve for \( \mu \):\[ (\mu - 1) = \frac{30 \times 6000 \times 10^{-10}}{3.6 \times 10^{-5}} \]\[ (\mu - 1) = \frac{180000 \times 10^{-10}}{3.6 \times 10^{-5}} \] \[ (\mu - 1) = 0.05 \]\[ \mu = 1.05 + 1\]\[ \mu = 1.2 \]

Step 6: Verify Result

The calculation shows that the refractive index \( \mu \) is 1.2. Thus, option (b) is the correct answer.

Key Concepts

Interference pattern

Interference pattern is a fascinating aspect of wave behavior, notably demonstrated in Young's Double Slit Experiment. When light passes through two closely spaced slits, it behaves like waves, creating a pattern of bright and dark bands on a screen. This pattern is known as an interference pattern and happens because the light waves overlap and combine.

• When the crests of two waves meet, they amplify each other, leading to a bright fringe or constructive interference.
• When a crest and a trough meet, they cancel each other out, resulting in a dark fringe or destructive interference.

The spacing and position of these fringes depend on the wavelength of the light and the distance between the slits. In the context of the problem, the interference pattern is disturbed by introducing a transparent sheet, which shifts fringe positions.

Refractive index

The refractive index, denoted by the Greek letter \(\mu\), is a measure of how much a material slows down light as it passes through. It's a crucial quantity in optics that tells us how light bends when it enters a different medium. The change in light speed is responsible for bending and is related to how dense or optically active a material is. When light encounters a transparent sheet, as in Young’s experiment, its speed changes based on the refractive index, creating a path difference between the two waves entering the slits. By determining this index, we can understand the optical properties of the material affecting the light. Refractive index is always more than or equal to 1, with 1 indicating no refractive change, like in a vacuum. In the problem, we calculated the refractive index to be 1.2, indicating that the light slows down by 20% in the sheet compared to air.

Path difference calculation

Path difference refers to the difference in the distance traveled by two coherent waves when reaching a particular point. It plays a crucial role in creating interference patterns. In Young's experiment, path difference is crucial because it determines the intensity of the fringes observed. Fringe shifts occur when additional path differences are introduced, such as by inserting a material with a different refractive index.To calculate the path difference introduced by a sheet of refractive index \(\mu\) and thickness \(t\), we use the formula:\[ \Delta x = (\mu - 1) t \]This formula helps in understanding how much the central fringe is displaced, as in the exercise where it ended up at the position of the 30th fringe.

• The thickness of the sheet, in this case, changes the path difference by creating optical length variations due to the refractive index.
• This calculated path difference matched the shift corresponding to 30 fringes.

Mastering this concept allows prediction of fringe shifts, a vital part of exploiting optical properties in experiments and practical applications.