Respuesta :
Answer:
Part A:
a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.
Part B:
b) If the spacing between the slits is decreased the fringe spacing Δy will increase.
Part C:
a) If the distance to the screen is decreased the fringe spacing will decrease.
Part D:
The dot in the center of fringe E is [tex]920\ x\ 10^{-9} m[/tex] farther from the left slit than from the right slit.
Explanation:
In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.
The position of bright fringes in the screen where the pattern is formed can be calculated with
[tex]\vartriangle y =\frac{m \lambda L}{d} [/tex]
[tex]m=0,\pm 1,\pm 2,\pm 3,.....[/tex]
- m is the order number.
- [tex]\lambda[/tex] is the wavelength of the monochromatic light.
- L is the distance between the screen and the two slits.
- d is the distance between the slits.
- Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light [tex]\lambda[/tex] is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
- Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
- Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
- Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be [tex]\vartriangle r=m \lambda[/tex]
We simply replace the values in that equation :
[tex]\vartriangle r= m \lambda =2.\ 460\ nm[/tex]
[tex]\vartriangle r= 920\ x\ 10^{-9} m[/tex]
The dot in the center of fringe E is [tex]920\ x\ 10^{-9}m[/tex] farther from the left slit than from the right slit.
By using the general equation for the double-slit maximums, we will get
- A) a
- B) b
- C) a
- D) Can't be done completely, but below there is a approach to it.
What is the distance between consecutive maximums?
The equation for the constructive interference in a double-slit experiment is given by:
[tex]y = \frac{m*\lambda*D}{d} [/tex]
Where:
- m is the number of the maximum.
- λ is the wavelength.
- D is the distance between the double-slit and the screen.
- d is the distance between the slits.
Now let's answer:
A) If the wavelength is decreased, then the numerator is decreased, meaning the separation between consecutive fringes will also be decreased, so the correct option is a.
B) If d is decreased then the denominator decreases, meaning that the distance between consecutive fringes increases, so the correct option is b.
C) is the distance D is decreased, similar like in case A, the numerator decreases, meaning that the correct option is a again.
D) Sadly, as we do not know:
- Which fringe is E.
- The value of D
- The value of d.
We can't answer this question.
What we should do here, is to compare the distance between the fringe and each slit, that distance will be the hypotenuse of a right triangle with one cathetus equal to D, and the other cathetus equal to y ± d/2, where each sign corresponds to each slit.
Then the difference in the distance will just be:
[tex]\sqrt{(y - d/2)^2 + D^2} - \sqrt{(y + d)^2 + D^2} [/tex]
If you want to learn more about the double-slit experiment, you can read:
https://brainly.com/question/13111431