To find the wavelength of light and the distance of the fourth bright fringe from the central bright fringe in a double-slit interference pattern, we can use the principles of wave interference. Let's break down the problem step by step.
Understanding the Setup
In a double-slit experiment, coherent light sources create an interference pattern on a screen. The distance between the slits (coherent sources) is given as 0.3 mm, and the distance from the slits to the screen is 90 cm. The second dark band is located 0.3 cm from the central bright fringe.
Key Formulas
We will use the following formulas related to interference patterns:
- The position of dark fringes is given by: y_d = (m + 0.5) * (λL / d)
- The position of bright fringes is given by: y_b = (m * λL / d)
Where:
- y_d = position of the dark fringe
- y_b = position of the bright fringe
- m = order of the fringe (0, 1, 2, ...)
- λ = wavelength of light
- L = distance from the slits to the screen
- d = distance between the slits
Finding the Wavelength
Given that the second dark band (m=1) is 0.3 cm from the central bright fringe, we can set up the equation for the first dark fringe:
0.3 cm = (1 + 0.5) * (λ * 90 cm / 0.03 cm)
Converting the distances to the same units, we have:
0.3 cm = 1.5 * (λ * 90 / 0.03)
Now, rearranging to solve for λ:
λ = (0.3 cm * 0.03 cm) / (1.5 * 90 cm)
Calculating this gives:
λ = (0.009 cm) / (135) = 0.00006667 cm = 6.667 x 10^-7 cm = 667 nm
Calculating the Distance of the Fourth Bright Fringe
Now that we have the wavelength, we can find the distance of the fourth bright fringe (m=4) from the central bright fringe:
Using the formula for bright fringes:
y_b = (4 * λ * L / d)
Substituting the values:
y_b = (4 * 667 nm * 90 cm / 0.03 cm)
First, convert the wavelength to cm:
667 nm = 6.67 x 10^-5 cm
Now substituting:
y_b = (4 * 6.67 x 10^-5 cm * 90 cm / 0.03 cm)
Calculating this gives:
y_b = (4 * 6.67 x 10^-5 * 90 / 0.03) = 0.000267 cm * 90 / 0.03 = 0.000267 cm * 3000 = 0.801 cm
Final Results
In summary, the wavelength of the light used in this experiment is approximately 667 nm, and the distance of the fourth bright fringe from the central bright fringe is about 0.801 cm.
