Solved: Width of Slit with Monochromatic Light 550nm is 0.055mm

  • Thread starter somecelxis
  • Start date
  • Tags
    Slit Width
In summary, a converging lens with a focal length of 50cm is used to focus a diffraction pattern of monochromatic light with a wavelength of 550nm onto a screen. The separation between the first minima is given as 10.0mm and the width of the slit is calculated to be 0.055mm using the formula x= λD/ a. However, there is uncertainty about whether the formula is meant to give the distance between the two first-order minima or between the central maximum and one of the first-order minima.
  • #1
somecelxis
121
0

Homework Statement


Monochromatic light of wavelength 550nm is used to illuminate normally a narrow slit. A converging lens of focal length 50cm is used to focus the diffraction pattern onto a screen. If the separation between the first minima is 10.0mm What is the width of the slit. The ans is 0.055mm.


Homework Equations





The Attempt at a Solution


by using formula x= λD/ a ,
i have 10x10^-3 = (550x10^-9) x ( 50x10^-2) / (a) ,
i get my a = 0.0275mm ... What's wrong with my working?
 
Physics news on Phys.org
  • #2
somecelxis said:
by using formula x= λD/ a ...

Is this formula meant to give the distance between the two first-order minima or is it meant to give the distance between the central maximum and one of the first-order minima?
 
  • #3
I guess the op means the distance between two first min is x ?
 
  • #4
In the working above I assume the x is distance between two first minimum... Anything wrong with it?
 
  • #5
x is the distance of any of the first minima from the central maximum.

ehild
 

1. What is the significance of the 550nm wavelength in this experiment?

The 550nm wavelength is significant because it falls within the visible light spectrum, making it suitable for human observation. It is also a common wavelength used in experiments involving monochromatic light.

2. How was the width of the slit determined in this experiment?

The width of the slit was determined by measuring the diffraction pattern produced by the monochromatic light passing through the slit. The distance between the first dark fringes on either side of the central bright fringe corresponds to the width of the slit.

3. What is the purpose of using monochromatic light in this experiment?

The use of monochromatic light ensures that only one specific wavelength is passing through the slit, allowing for more accurate measurements of the diffraction pattern. This eliminates any potential interference from other wavelengths of light.

4. How does changing the width of the slit affect the diffraction pattern?

As the width of the slit decreases, the diffraction pattern widens and becomes more spread out. This is because a narrower slit allows for more diffraction to occur, resulting in more interference patterns and a wider diffraction pattern.

5. Can this experiment be replicated with different wavelengths of light?

Yes, this experiment can be replicated with different wavelengths of light. However, the width of the slit may need to be adjusted depending on the wavelength used in order to obtain accurate results and a visible diffraction pattern.

Similar threads

Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
4K
  • Introductory Physics Homework Help
Replies
34
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
5K
  • Introductory Physics Homework Help
Replies
10
Views
8K
Back
Top