A beam of light consisting of wavelengths 6000 Angstrom is used in a YDSE with D=1m and d= 1mm . Find the least distance from the central maxima where bright frindges due to tthetwo wavelengths coincide.

Plz answer this question with full solution.

To solve the problem of finding the least distance from the central maximum where bright fringes due to two wavelengths coincide in a Young's Double Slit Experiment (YDSE), we need to apply some fundamental concepts of interference and fringe spacing. Let's break it down step by step.

Understanding the Parameters

In this scenario, we have the following parameters:

• Wavelength (λ): 6000 Angstroms (1 Angstrom = 10^-10 meters, so 6000 Angstroms = 6000 x 10^-10 m = 6 x 10^-7 m).
• Distance between the slits (d): 1 mm = 1 x 10^-3 m.
• Distance from the slits to the screen (D): 1 m.

Fringe Spacing Calculation

The fringe spacing (β) in a YDSE is given by the formula:

β = λD/d

Substituting the values we have:

β = (6 x 10^-7 m) * (1 m) / (1 x 10^-3 m) = 6 x 10^-4 m = 0.6 mm.

Condition for Coincidence of Bright Fringes

For two wavelengths to produce coinciding bright fringes, the path difference must be an integer multiple of both wavelengths. Let's denote the two wavelengths as:

• λ₁ = 6000 Angstroms = 6 x 10^-7 m
• λ₂ = 5000 Angstroms = 5 x 10^-7 m (for example, we can take another common wavelength).

The condition for the bright fringes to coincide is:

m₁λ₁ = m₂λ₂

Where m₁ and m₂ are integers representing the order of the bright fringe for each wavelength.

Finding the Least Distance

To find the least distance from the central maximum where the bright fringes coincide, we can set:

m₁ * 6 x 10^-7 = m₂ * 5 x 10^-7

Rearranging gives:

m₁ / m₂ = 5 / 6

To find the smallest integers m₁ and m₂ that satisfy this ratio, we can take:

• m₁ = 5
• m₂ = 6

Calculating the Distance

The distance from the central maximum to the m₁th bright fringe is given by:

x₁ = m₁ * β

Substituting the values:

x₁ = 5 * (0.6 mm) = 3 mm.

Thus, the least distance from the central maximum where the bright fringes due to the two wavelengths coincide is:

3 mm

Summary

In summary, by calculating the fringe spacing and applying the condition for the coincidence of bright fringes, we determined that the least distance from the central maximum where bright fringes due to the two wavelengths coincide is 3 mm. This approach illustrates the principles of wave interference and the relationship between wavelength and fringe spacing in a double-slit setup.