Solving 2 Slit Diffraction: Find Central & 10th Maxima Intensity

In summary, a HeNe laser beam with a diameter of 1cm and a wavelength of 634nm illuminates a pair of 0.1mm wide slits with a separation of 2mm. An interference pattern is formed on a screen 1m away. The central maximum and 1st, 2nd, and 10th maxima can be found using the formula sinθ = (mλ)/d. The intensity of the 10th maximum can be determined using the formula I(θ) = Im(cos²β)(sinα/α)² and the index of refraction of CO2 at STP is 1.000449. When the room is filled with CO2, the
  • #1
stunner5000pt
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A coherent, monochromatic HeNe laser beam 1cm in diameter at lambda = 634nm illuminates a pair of slits. The separation of the slits is 2mm and the slits are each 0.1mm across. An interference pattern is cast on a screen 1m away from the slits. Index of refraction of air = 1.0002926 and CO2 (at STP) = 1.000449.

Find the location of the central maximimum and the 1st 2nd and 10th maxima
Now normally this isn't a big deal all i use is the formula
[tex] sin \theta = \frac{m \lambda}{d} [/tex] and [tex] \frac{\Delta x}{L} = \frac{\lambda}{d} [/tex]
d = 2.1mm
lambda = 634nm
L = 1m
the thing which throws me off is the diameter of the beam. Does that have any effect on the way this pattern appears??

b) Find the intensity of the 10 th maximum compared to the intensity of the central maximum

using this formula [tex] I(\theta) = I_{m} (cos^2 \beta) (\frac{sin \alpha}{\alpha})^2 [/tex]
where [tex] \alpha = \frac{\pi a}{\lambda} sin \theta [/tex]
[tex] \beta = \frac{\pi a}{\lambda} sin \theta [/tex]
and a = width of slit = 0.1mm
for the central max the angle theta is zero and for the 10th the angle is determined from above

c) Now the room is filled with CO2 uniformly. What is the index of refraction of the Co2 gas? In three sentence or less explain how the intensity of the pattern will change

well since carbon di oxide has a different index of refraction (which is 1.000449)
should i use this little relation
[tex] \frac{n_{air}}{n_{CO_{2}}} = \frac{v_{CO_{2}}}{v_{air}} [/tex]
and this is also the reation of the wavelengths and the frequencies so equate them all

Please tell me if I am wrong, i quite stumped on the width of the beam... How would i deal with that??
 
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can anyone help??

my only problem is what to do width the width of the beam?? Any help would be greatly appreciated!
 
  • #3


The width of the beam will not have a significant effect on the interference pattern. It will only slightly broaden the peaks and troughs of the pattern, but the overall location and intensity of the maxima will not change significantly.

a) Using the formula for the location of maxima, we can find the central maximum at \theta = 0. We can then plug in values for m (1, 2, and 10) to find the angle \theta for the 1st, 2nd, and 10th maxima. The corresponding distances from the central maximum can then be found using the formula \frac{\Delta x}{L} = \frac{\lambda}{d}.

b) To find the intensity of the 10th maximum compared to the central maximum, we can use the formula I(\theta) = I_{m} (cos^2 \beta) (\frac{sin \alpha}{\alpha})^2, where \beta and \alpha are determined using the values for a, \lambda, and \theta for the 10th maximum. The intensity of the 10th maximum will be significantly lower than the central maximum due to the decrease in intensity with increasing distance from the central maximum.

c) The index of refraction for CO2 at STP is 1.000449, which is slightly higher than that of air (1.0002926). As a result, the wavelength of the laser beam will decrease slightly in the presence of CO2, causing the interference pattern to shift slightly and the intensity to decrease. This change in the index of refraction will have a small effect on the interference pattern, but it will not significantly change the overall appearance.
 

1. What is the 2-slit diffraction phenomenon?

The 2-slit diffraction phenomenon is a behavior of light or other waves passing through two closely spaced slits or obstacles. It results in a pattern of alternating bright and dark fringes on a screen behind the slits, known as an interference pattern. This phenomenon is a fundamental concept in the field of wave optics and is used to study the properties of light and other types of waves.

2. What is the central maximum in a 2-slit diffraction pattern?

The central maximum is the central bright fringe in a 2-slit diffraction pattern. It is the brightest point on the screen and is directly in line with the slits. This maximum occurs because the waves from each slit interfere constructively at this point, resulting in a high intensity of light.

3. How is the central maximum intensity calculated?

The central maximum intensity is calculated using the equation I = I0(sin(θ)/θ)^2, where I0 is the intensity of the incident light, θ is the angle between the central maximum and the first dark fringe, and I is the intensity at a given point on the screen. This equation is derived from the wave theory of light and is used to determine the intensity at any point in the diffraction pattern.

4. What is the 10th maximum in a 2-slit diffraction pattern?

The 10th maximum is the 10th bright fringe in the diffraction pattern, counted from the central maximum. It is also known as the 10th order maximum. As the order of the maximum increases, the intensity of the fringe decreases, but the spacing between fringes remains the same.

5. How is the position of the 10th maximum calculated?

The position of the 10th maximum can be calculated using the equation y10 = 10λD/d, where y10 is the distance from the central maximum to the 10th maximum on the screen, λ is the wavelength of the incident light, D is the distance from the slits to the screen, and d is the distance between the slits. This equation is based on the principle of constructive interference and is used to determine the position of any order maximum in the diffraction pattern.

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