Interference and diffraction of light

In summary, using the equation wavelength=d sin degree/ n, the spacing between lines on the grating can be calculated by solving for d. Once d is found, it can be converted into centimeters to determine the number of lines per centimeter.
  • #1
poohead
34
0

Homework Statement


green light of wavelength 5000 A is incident normally on a grating, and the 2nd order image is diffracted 32 degrees from the normal. how many lines/cm are marked on the grating

A=10^-10 n= 2 degree= 32 d=? wavelength- 5000 A

the answer to the question is 5300 lines/cm

Homework Equations



wavelength=d sin degree/ n

The Attempt at a Solution



we tried to solve for d using the following above equation but came out with the wrong answer, need help please i am in a fatal situation here,

 
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  • #2
poohead said:
we tried to solve for d using the following above equation but came out with the wrong answer, need help please i am in a fatal situation here,
d is the distance between slits, not the number of slits per cm.
 
  • #3
So then what would I do to get an answer with lines/cm
 
  • #4
poohead said:
So then what would I do to get an answer with lines/cm
Given the spacing between lines, figure out how many fit into a centimeter.
 
  • #5
the spacing between lines is X?
 
  • #6
I don't understand then what equation i am to use to solve for such
 
  • #7
poohead said:
I don't understand then what equation i am to use to solve for such
Here's an example: If d = 1/100 of a cm, how many lines are there per centimeter?
 
  • #8
100? but you said that d is the distance between slits, not the number of slits per cm. so what am i trying to find for this situation?
 
  • #9
poohead said:
100?
Right. If you can do this example, use similar reasoning to solve your problem.
but you said that d is the distance between slits, not the number of slits per cm.
That's true. You use d to calculate the number of slits per cm.
so what am i trying to find for this situation?
Your final answer will be the number of lines per cm. You start by finding d.
 
  • #10
sorry I am still confused on how to get there, what would i possibly do with d?
 
  • #11
poohead said:
sorry I am still confused on how to get there, what would i possibly do with d?
Try this analogy: Along a path there is a tree every 1/4 mile. How many trees per mile?

Here "1/4 mile" plays the same role as "d".
 
  • #12
there would be four, but so i solved for d and i get 1.81x10^-6, so now i convert that into cm?
 
  • #13
poohead said:
so i solved for d and i get 1.81x10^-6,
Careful how you round off.
so now i convert that into cm?
Yes, find d in cm.
 

Related to Interference and diffraction of light

1. What is interference and diffraction of light?

Interference and diffraction of light are phenomena that occur when light waves interact with each other or with obstacles, causing changes in the intensity and direction of the light. These phenomena are important in understanding the behavior of light and its properties.

2. What is the difference between interference and diffraction?

Interference occurs when two or more light waves combine and either reinforce or cancel each other out, resulting in a pattern of light and dark fringes. Diffraction, on the other hand, refers to the bending of light waves around obstacles or through small openings, resulting in the spreading out of the light.

3. What is the relationship between wavelength and interference/diffraction?

The behavior of interference and diffraction is dependent on the wavelength of light. Shorter wavelengths, such as blue light, experience more diffraction, while longer wavelengths, such as red light, experience less diffraction. Interference, on the other hand, is dependent on the spacing between the light sources or obstacles.

4. How is interference and diffraction of light used in practical applications?

Interference and diffraction of light have many practical applications in various fields. In optics, they are used to create diffraction gratings, which are used in spectroscopy to separate light into its component wavelengths. In technology, they are used in devices such as holograms and anti-reflective coatings. They also play a role in medical imaging, astronomy, and telecommunications.

5. What are some real-life examples of interference and diffraction of light?

Interference and diffraction of light can be observed in everyday life. For example, the colors seen on a soap bubble or an oil slick are a result of light waves interfering with each other. The rainbow effect seen on a CD or DVD is due to the diffraction of light off the disc's surface. Sunsets and sunrises also exhibit diffraction, as the longer wavelengths of light are scattered more by the Earth's atmosphere, creating a red or orange hue.

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