# Double slit: ratio of intensity of 3rd- and 0th-order maxima

• DottZakapa
In summary, the goal of the homework statement is to find the ratio of the intensity of the third order maximum with that of the zero-order maximum. The equations for interference diffraction tell the student that the intensity of the third order maximum is equal to the intensity of the zero-order maximum multiplied by cos2(β). The student then finds the angle θ between the slits and solves for β. Finally, the student finds the distance between the slits, D, and the wavelength, λ, using the equation d = 6λ.
DottZakapa
Homework Statement
In a double slit experiment let d=5.00 D=30.0λ. Estimate the ratio of the intensity of the third order maximum with that of the zero-order maximum.
Relevant Equations
interference diffraction
Homework Statement: In a double slit experiment let d=5.00 D=30.0λ. Estimate the ratio of the intensity of the third order maximum with that of the zero-order maximum.
Homework Equations: interference diffraction

i guess the goal is this equation

##I_{(\theta)}=I_0 \times(cos^2\beta)\times \left ( \frac {sin\alpha} \alpha \right)^2##

then i do

## D\sin \theta = 3\lambda##

##\sin\theta= \frac {3\lambda} D, \space \theta=5.74^0##

##\beta= \frac {\pi d} \lambda \sin \theta##

##\alpha = \frac {\pi D} \lambda \sin \theta \space##

substituting the data

##\alpha=\frac {\pi 30.0\lambda} \lambda \frac {3\lambda} {30.0\lambda}\space##
next

##\beta= \frac {\pi 5.00} \lambda \frac {3\lambda} {30.0\lambda}## i don't know how to solve this one, and solve the rest of the problem, how do i get rid of ##\space\lambda\space## at denominator?

DottZakapa said:
Homework Statement: In a double slit experiment let d=5.00 D=30.0λ. Estimate...
d is 5.00 what? Meters, inches, light years, ...

kuruman said:
d is 5.00 what? Meters, inches, light years, ...
That is what the text says. There is no other data informato. But i guess that d is the distance between the two slits
And D is the slits width.

DottZakapa said:
That is what the text says. There is no other data informato. But i guess that d is the distance between the two slits
And D is the slits width.
I'm sure you are right about the meaning of d and D. Nevertheless, units are needed for d. Note that if you can express d as a multiple of the wavelength λ, then you will be able to get rid of the extra λ in the denominator in the expression for ##\beta##.

kuruman said:
I'm sure you are right about the meaning of d and D. Nevertheless, units are needed for d. Note that if you can express d as a multiple of the wavelength λ, then you will be able to get rid of the extra λ in the denominator in the expression for ##\beta##.
this is the solution but... i don't understand from where those result come from

no one knows?

Looks like the solution is using D = 6λ and d = 30λ. (These are different values from what you gave in the statement of the problem.) Also, it appears to me that there is a mistake in the formula for ##I## given in the solution. I think the factor of ##\sin^2(\delta/2)## should be ##\cos^2(\delta/2)##.

TSny said:
I think the factor of ##\sin^2(\delta/2)## should be ##\cos^2(\delta/2)##.
you right, unfortunately i did not have the chance to meet the teacher and ask

## 1. What is the double slit experiment?

The double slit experiment is a classic experiment in physics that demonstrates the wave-particle duality of light. It involves shining a beam of light through two parallel slits and observing the resulting interference pattern on a screen.

## 2. What is the ratio of intensity of the 3rd- and 0th-order maxima in the double slit experiment?

The ratio of intensity of the 3rd- and 0th-order maxima in the double slit experiment depends on the distance between the two slits, the wavelength of the light, and the distance from the slits to the screen. This ratio can be calculated using the equations of diffraction and interference.

## 3. How does the distance between the two slits affect the ratio of intensity of the 3rd- and 0th-order maxima?

The distance between the two slits directly affects the ratio of intensity of the 3rd- and 0th-order maxima in the double slit experiment. As the distance between the slits increases, the ratio of the intensities decreases, resulting in a wider central maximum and narrower secondary maxima.

## 4. Why does the ratio of intensity of the 3rd- and 0th-order maxima change with different wavelengths of light?

The ratio of intensity of the 3rd- and 0th-order maxima changes with different wavelengths of light because different wavelengths have different diffraction and interference patterns. This means that the maxima and minima will occur at different points on the screen, resulting in a different ratio of intensities.

## 5. How is the double slit experiment used in modern science?

The double slit experiment is still used in modern science to study the behavior of light and other particles. It has also been adapted to study the wave-particle duality of other particles, such as electrons and atoms. The experiment has also been used to test the limits of quantum mechanics and to develop new technologies, such as the electron microscope.

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