To find the distance \( x \) above the central maximum where the intensity is 75% of the maximum in a double slit experiment, we can use the principles of interference and the formula for intensity in terms of the path difference. Let's break this down step by step.
Understanding the Double Slit Experiment
In a double slit experiment, coherent light passes through two closely spaced slits, creating an interference pattern on a screen. The pattern consists of alternating bright and dark fringes due to constructive and destructive interference of light waves.
Key Variables
- d: Distance between the slits = 0.250 cm = 0.00250 m
- R: Distance from the slits to the screen = 120 cm = 1.20 m
- λ: Wavelength of light = 600 nm = 600 x 10-9 m
Intensity Formula
The intensity \( I \) at a point on the screen can be expressed in terms of the maximum intensity \( I_0 \) and the path difference \( \Delta x \) as follows:
I = I_0 \cos^2\left(\frac{\pi \Delta x}{\lambda}\right)
To find the distance \( x \) where the intensity is 75% of the maximum intensity, we set up the equation:
I = 0.75 I_0
Substituting this into the intensity formula gives:
0.75 I_0 = I_0 \cos^2\left(\frac{\pi \Delta x}{\lambda}\right)
Dividing both sides by \( I_0 \) leads to:
0.75 = \cos^2\left(\frac{\pi \Delta x}{\lambda}\right)
Finding the Path Difference
The path difference \( \Delta x \) for a point at distance \( x \) from the central maximum is given by:
\(\Delta x = \frac{d \cdot x}{R}\)
Substituting this into the equation gives:
0.75 = \cos^2\left(\frac{\pi d x}{\lambda R}\right)
Solving for \( x \)
Taking the square root of both sides, we have:
\(\sqrt{0.75} = \cos\left(\frac{\pi d x}{\lambda R}\right)\)
Calculating \( \sqrt{0.75} \) gives approximately 0.866. Thus, we have:
0.866 = \cos\left(\frac{\pi d x}{\lambda R}\right)
Now, we can find the angle whose cosine is 0.866:
\(\theta = \cos^{-1}(0.866) \approx 30^\circ\)
Calculating \( x \)
Using the small angle approximation, where \( \tan(\theta) \approx \sin(\theta) \approx \theta \) in radians, we can relate \( x \) to \( \theta \):
\(\frac{d x}{R} \approx \sin(\theta)\)
Substituting the values:
\(\frac{0.00250 \cdot x}{1.20} \approx \sin(30^\circ) = 0.5\)
Now, solving for \( x \):
x = \frac{0.5 \cdot 1.20}{0.00250} = 240 m
Final Result
Thus, the distance \( x \) above the central maximum where the intensity is 75% of the maximum is approximately 0.240 m or 24.0 cm. This calculation illustrates how interference patterns can be quantitatively analyzed using the principles of wave optics.
