About Newtion ring experiment

In summary, for the given problem of finding the distance d using the refractive index of a prism, a wavelength of 450nm and 50 observed bright rings, the condition for constructive interference is that the path difference is equal to the wavelength, or 2d=nλ/2. However, the correct condition may be that the path difference is equal to the wavelength plus half a wavelength, or 2d+(λ/2)=nλ. It is recommended to check a textbook for the correct condition and to draw a ray diagram for a better understanding of interference of two waves.
  • #1
58
1

Homework Statement


擷取3.PNG

refractive index of prism=1.45
λ=450nm
50 bright ring are observed
find d

Homework Equations


2d=nλ/2

The Attempt at a Solution


since 50 bright rings are observed n=24 (-24---------0-------------24)
if the distance travel=λ/2 constructive interference will happen
since the ray1 reflected by prism(n=1.45--->n=1 air)no phase shift
by ray 2 air to flat plate have phase shift

2d=24(λ/2) and λ=450nm/n=450nm/1.45=310nm
d=[310nm(24)]/4=1.86x10^-6m but the answer=1.11x10^5m why?
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  • #2
kenok1216 said:

Homework Equations


2d=nλ/2
i think
for constructive interference -draw a diagram of the rays interfering at thickness d
the correct condition perhaps is
path difference = n. wavelength (in phase)
and path difference =2.d + (wavelength / 2)
check from your textbook
 
  • #3
drvrm said:
i think
for constructive interference -draw a diagram of the rays interfering at thickness d
the correct condition perhaps is
path difference = n. wavelength (in phase)
and path difference =2.d + (wavelength / 2)
check from your textbook
so do you mean nλ=2d+
drvrm said:
i think
for constructive interference -draw a diagram of the rays interfering at thickness d
the correct condition perhaps is
path difference = n. wavelength (in phase)
and path difference =2.d + (wavelength / 2)
check from your textbook
why pd=2d+λ/2?
 
  • #4
kenok1216 said:
so do you mean nλ=2d+

why pd=2d+λ/2?
draw the ray diagram for interference
 

1. What is the purpose of the Newton ring experiment?

The Newton ring experiment is used to measure the wavelength of light. It is also used to study the phenomenon of interference of light waves.

2. How does the Newton ring experiment work?

The experiment involves placing a convex lens on top of a flat glass plate, with a thin layer of air in between. When light is shone on this setup, the air layer acts as a wedge, causing interference patterns known as Newton's rings to form. These patterns can be used to calculate the wavelength of the light.

3. What are the factors that affect the results of the Newton ring experiment?

The main factors that affect the results of the experiment are the thickness of the air layer, the curvature of the lens, and the wavelength of the light being used.

4. What are some practical applications of the Newton ring experiment?

The Newton ring experiment is commonly used in research and industry to measure the refractive index of materials, as well as to calibrate optical instruments. It is also used in the production of high precision lenses and mirrors.

5. Are there any limitations to the Newton ring experiment?

Yes, there are a few limitations to the experiment. It requires very precise measurements and can be affected by external factors such as air currents. It also only works with monochromatic light, so it cannot be used to measure the refractive index of materials for all wavelengths of light.

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