# HW help (Double Slit, Resolving power etc)

• Alt+F4
In summary, The conversation discusses the setup and conditions for a double-slit experiment using red light and the conditions for constructive and destructive interference. The second part of the conversation talks about a new experiment with different wavelengths and distances, and the process of finding the distance between different order bright fringes. Finally, the conversation ends with confirming the correct answer for the second part and asking for the answer to the first part.
Alt+F4
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(a) A double-slit experiment is set up using red light (l = 708 nm). A first order bright fringe is seen at a given location on a screen. What wavelength of visible light (between 380 nm and 750 nm) would produce a dark fringe at the identical location on the screen?
l = nm

All right so i know
d * sin theta = m * Wavelenght

How do i find the Distance and the angle?

The condition for constructive interference is what you have written down (i.e. peak on peak), but destructive interference happens when you have a trough meet a peak, i.e. there is a half wavelength difference so for destructive interference (dark fringe) $$dsin (\theta) = (m + 0.5)\lambda$$. You should be able to go from there.

(b) A new experiment is created with the screen at a distance of 1.8 m from the slits (with spacing 0.08 mm). What is the distance between the second order bright fringe of light with l = 691 nm and the third order bright fringe of light with l = 414 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.)

Ok so
D sin theta = m * Wavelenght

Distance is 2.2?
I am going to find theta
m = 1
Wavelenght is 698 in one part, and then i do the quation again for 413 right?

Alt+F4 said:
(b) A new experiment is created with the screen at a distance of 1.8 m from the slits (with spacing 0.08 mm). What is the distance between the second order bright fringe of light with l = 691 nm and the third order bright fringe of light with l = 414 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.)

Ok so
D sin theta = m * Wavelenght

Distance is 2.2?
I am going to find theta
m = 1
Wavelenght is 698 in one part, and then i do the quation again for 413 right?
Don't lose track of the statement about the different orders of the finges. What is m for second order, and for third order?

OlderDan said:
Don't lose track of the statement about the different orders of the finges. What is m for second order, and for third order?
2 for second order, and 3 for third order

Last edited:
You don't actually need to find the distance at all. You will have two equations equal to $$dsin (\theta)$$, and you just have to solve for the 1 unknown (wavelength).

Alt+F4 said:

and that would be?

.00315 m
...

Yes that's correct for (b). What answer did you get for (a).

## 1. What is the double slit experiment and why is it important?

The double slit experiment is a classic physics experiment that involves shining a beam of light through two parallel slits and observing the interference pattern that is created on a screen behind the slits. It is important because it demonstrates the wave-like nature of light and supports the concept of superposition, where two waves can combine to create a new wave.

## 2. How is the resolving power of a telescope calculated?

The resolving power of a telescope is calculated by dividing the wavelength of light being observed by the diameter of the telescope's objective lens or mirror. This calculation gives the smallest angle at which two closely spaced objects can be distinguished by the telescope.

## 3. What factors affect the resolving power of a telescope?

The resolving power of a telescope can be affected by several factors, including the diameter of the objective lens or mirror, the quality of the optics, the atmospheric conditions, and the wavelength of the light being observed.

## 4. How does the double slit experiment relate to the resolving power of telescopes?

The double slit experiment demonstrates the diffraction of light, which is also a factor in the resolving power of telescopes. The smaller the aperture (or slit) through which light passes, the greater the diffraction and the lower the resolving power of the telescope.

## 5. Can the resolving power of a telescope be improved?

Yes, the resolving power of a telescope can be improved by increasing the diameter of the objective lens or mirror, using higher quality optics, and observing in better atmospheric conditions. For ground-based telescopes, adaptive optics can also be used to correct for distortions caused by atmospheric turbulence, improving the resolving power even further.

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