In a Young's double slit experiment, we can determine which wavelength is absent at a specific point on the screen by using the concept of interference. The key idea here is that constructive and destructive interference occur based on the path difference between the light waves coming from the two slits. Let's break this down step by step.
Understanding the Setup
In this experiment, light passes through two closely spaced slits and creates an interference pattern on a screen. The distance between the slits is denoted as d, and the distance from the slits to the screen is D. The wavelengths of light range from 420 nm to 700 nm.
Path Difference and Interference
The condition for destructive interference (where a wavelength is absent) at a point on the screen is given by the formula:
- Path Difference = (m + 0.5) * λ
Here, m is an integer (0, 1, 2, ...), and λ is the wavelength of the light. The point where we are checking for the absence of a wavelength is 1 mm away from the central fringe.
Calculating the Path Difference
The path difference at a distance y from the central maximum can be calculated using the formula:
- Path Difference = (d * y) / D
Substituting the values:
- d = 0.5 mm = 0.5 x 10-3 m
- y = 1 mm = 1 x 10-3 m
- D = 50 cm = 0.5 m
Now, we can calculate the path difference:
Path Difference = (0.5 x 10-3 m * 1 x 10-3 m) / 0.5 m = 1 x 10-6 m = 1000 nm
Finding the Wavelengths Absent
Next, we need to find which of the given wavelengths satisfy the condition for destructive interference:
- 500 nm: 1000 nm = (1 + 0.5) * 500 nm (m = 1) - present
- 667 nm: 1000 nm = (1 + 0.5) * 667 nm (m = 1) - present
- 600 nm: 1000 nm = (1 + 0.5) * 600 nm (m = 1) - present
Since all calculated wavelengths lead to constructive interference, we need to check for the absence of wavelengths that would lead to destructive interference at this path difference. The only wavelength that would satisfy the condition for being absent is:
- All of these wavelengths can be present at the point 1 mm away from the central fringe.
Final Thoughts
In conclusion, since the path difference of 1000 nm does not correspond to any of the wavelengths provided in a way that would cause them to be absent, the answer to the question is that none of the wavelengths are completely absent at that point. Thus, the correct choice is:
