In a Young’s double-slit experiment, light behaves as a wave, and when it passes through two closely spaced slits, it creates an interference pattern on a screen. The bright fringes occur at specific distances from the central maximum, and this distance can be calculated based on the wavelength of the light used. Let's break down your question step by step.
Finding the Distance of the Third Bright Fringe
To calculate the position of the bright fringes, we can use the formula:
y = (mλL) / d
Where:
- y = distance from the central maximum to the m-th bright fringe
- m = order of the fringe (for the third bright fringe, m = 3)
- λ = wavelength of light (in meters)
- L = distance from the slits to the screen (in meters)
- d = distance between the slits (in meters)
For wavelength 650 nm, we need to convert the wavelength into meters:
λ = 650 nm = 650 × 10-9 m
Assuming we know the values for L and d, we can now calculate the position of the third bright fringe. Let's say L is 1 meter and d is 0.0005 meters (these are just example values, you can replace them with actual measurements if available).
Plugging in the values:
y = (3 × 650 × 10-9 m × 1 m) / 0.0005 m
Calculating this gives:
y ≈ 0.0039 m or 3.9 mm
Determining the Coinciding Bright Fringes
Next, we want to find the least distance from the central maximum where the bright fringes of both wavelengths coincide. This occurs when the positions of the bright fringes for both wavelengths are equal. For the m-th bright fringe of both wavelengths, we set their respective equations equal:
(m₁λ₁L) / d = (m₂λ₂L) / d
Since L and d are common to both, they can be eliminated:
m₁λ₁ = m₂λ₂
Now we can substitute the wavelengths:
m₁(650 nm) = m₂(520 nm)
To find the smallest values of m₁ and m₂ that satisfy this equation, we can express the relationship as:
m₁ / m₂ = 520 / 650 = 0.8
This indicates that the smallest integer values that satisfy the equation are m₁ = 8 and m₂ = 10. Now we can find the distance for either wavelength:
Using m₁ = 8 for 650 nm:
y = (8 × 650 × 10-9 m × L) / d
Similarly, you can use the respective values for m₂ and 520 nm to get the same y value. This will give you the least distance from the central maximum where the bright fringes coincide. If we assume L = 1 m and d = 0.0005 m again:
y = (8 × 650 × 10-9 m × 1 m) / 0.0005 m
Thus, y ≈ 0.0104 m or 10.4 mm.
In summary, the distance of the third bright fringe for the wavelength of 650 nm is approximately 3.9 mm, and the least distance from the central maximum where the bright fringes due to both wavelengths coincide is approximately 10.4 mm. If you have different values for L or d, just substitute them into the formulas to get your final answers.
