Problem 66 - Master Resource Book in JEE Main Physics

Question

Two slits, \(4 \mathrm{~mm}\) apart are illuminated by light of wavelength \(600 \AA\). What will be the fringe width on a screen placed \(2 \mathrm{~m}\) from the slits? (a) \(0.12 \mathrm{~mm}\) (b) \(0.3 \mathrm{~mm}\) (c) \(3.0 \mathrm{~mm}\) (d) \(4.0 \mathrm{~mm}\)

Step-by-Step Solution

Answer

The fringe width is \(0.3 \text{ mm}\), which corresponds to option (b).

Step 1: Convert Wavelength to Meters

We need to convert the wavelength from Angstroms to meters to use it in our formula. The conversion is given by: \(1 \text{ Å} = 10^{-10} \text{ m}\). Thus, \(600 \text{ Å} = 600 \times 10^{-10} \text{ m} = 6 \times 10^{-8} \text{ m} \).

Step 2: Recall Fringe Width Formula

The formula for fringe width \( \beta \) in a double-slit experiment is given by: \[ \beta = \frac{\lambda \cdot D}{d} \]where \(\lambda\) is the wavelength of light, \(D\) is the distance from slits to screen, and \(d\) is the separation between slits.

Step 3: Substitute Known Values

Substituting the known values into the formula:\( \lambda = 6 \times 10^{-8} \text{ m}, D = 2 \text{ m}, d = 4 \times 10^{-3} \text{ m}\):\[ \beta = \frac{6 \times 10^{-8} \times 2}{4 \times 10^{-3}} \].

Step 4: Simplify the Calculation

Simplify the expression:\[ \beta = \frac{1.2 \times 10^{-7}}{4 \times 10^{-3}} = 0.3 \times 10^{-4} \text{ m} = 0.003 \text{ m} = 0.3 \text{ mm} \].

Key Concepts

Fringe Width

In Young's Double Slit Experiment, fringe width is a fascinating concept. It essentially refers to the distance between successive bright or dark interference fringes on a screen.

Understanding this helps us not only grasp the phenomenon of interference but also provides a quantitative measure of it. The fringe width can be calculated using the formula: \[ \beta = \frac{\lambda \cdot D}{d} \] where:

• \( \beta \) = fringe width
• \( \lambda \) = wavelength of the light used
• \( D \) = distance from the slits to the screen
• \( d \) = distance between the slits

The equation tells us how the fringe width (\(\beta\)) changes based on factors like the wavelength and slit distance. Larger fringes are produced by light of longer wavelengths, closer slits, or farther screen distances.
This intricate relationship helps us understand how light behaves under varying experimental setups.

Wavelength Conversion

Wavelength conversion is an essential step in many physics calculations, as light wavelengths can be expressed in different units. In this experiment, the light's wavelength is initially provided in Angstroms (\( \text{Å} \) ). To correctly apply it in our formula for fringe width, we need to convert it into meters (\( \text{m} \) ), the standard SI unit for length.

An Angstrom represents \(10^{-10} \text{m}\). To convert a wavelength, such as \(600 \text{ Å}\), into meters, you multiply it by \(10^{-10}\). Therefore, in the exercise, it becomes \( 6 \times 10^{-8} \text{m} \).
Proper conversion ensures calculations are accurate and consistent within the scientific community. Without this step, the results derived would be incorrect, highlighting the importance of unit consistency in physics.

Optical Interference

Optical interference forms the backbone of the Young's Double Slit Experiment. It's a phenomenon where waves overlap, causing them to either strengthen or weaken depending on their relative phases.

In this experiment, light waves from two slits interfere with one another creating a series of light and dark patterns on a screen.

• Constructive interference occurs when the wave peaks align, forming bright fringes.
• Destructive interference happens when the peaks and troughs align oppositely, resulting in dark fringes.

Understanding optical interference not only helps us appreciate this fundamental property of waves but is also crucial for technologies relying on the manipulation of light. It’s fascinating how these simple patterns result from complex behavior inherently present in wave dynamics.