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General Physics

In a Young’s double-slit experiment, the slits are separated by0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Profile image of samrat
12 Years agoGrade
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1 Answer

Profile image of Saurabh Koranglekar
6 Years ago

To determine the wavelength of light used in a Young's double-slit experiment, we can utilize the formula that relates the wavelength to the distance between the slits, the distance to the screen, and the positions of the fringes. Let's break this down step by step.

Understanding the Setup

In a typical Young's double-slit experiment, light passes through two closely spaced slits and creates an interference pattern on a screen. The distance between the central bright fringe (the zeroth order maximum) and any bright fringe can be described using the following relationship:

The Formula

The formula to calculate the position of the bright fringes is given by:

y = (m * λ * L) / d

  • y = distance from the central maximum to the m-th bright fringe
  • m = order of the bright fringe (0 for central, 1 for first order, 2 for second, etc.)
  • λ = wavelength of light
  • L = distance from the slits to the screen
  • d = distance between the two slits

Now, in your case:

  • Distance between the slits, d = 0.28 mm = 0.28 x 10^-3 m
  • Distance to screen, L = 1.4 m
  • Distance from the central maximum to the fourth bright fringe (m = 4), y = 1.2 cm = 0.012 m

Plugging in the Values

Now we can rearrange the formula to solve for the wavelength (λ):

λ = (y * d) / (m * L)

Substituting in the known values:

λ = (0.012 m * 0.28 x 10^-3 m) / (4 * 1.4 m)

Calculating the Wavelength

Let's do the math step-by-step:

  • Calculate the numerator: 0.012 m * 0.28 x 10^-3 m = 3.36 x 10^-6 m²
  • Calculate the denominator: 4 * 1.4 m = 5.6 m

Now, substituting these values back into the equation for wavelength:

λ = (3.36 x 10^-6 m²) / (5.6 m)

Calculating this gives:

λ ≈ 6.0 x 10^-7 m

Final Result

Thus, the wavelength of light used in this experiment is approximately 600 nm (nanometers), which falls within the visible spectrum, specifically in the orange-red light range.

Real-World Significance

This experiment not only demonstrates the wave nature of light but also provides a practical application of interference patterns, which are essential in various fields such as optics, telecommunications, and even quantum mechanics.