2. Green light with a frequency of 6 x 10¹4 Hz is shined on two slits separated a distance of 5 x 10-5 m. The detector screen is located 1.5 m away. a) Treating the slits as point-like, how far above the center of the detector is the first dark fringe? How about the first bright fringe (excluding the bright fringe at the center)? b) Consider the second dark fringe above the center of the detector and the first bright spot below the center of the detector, how far apart are these locations? c) Would your answer to a) increase, decrease, or stay the same if the light was violet? What if it was red?

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2. Green light with a frequency of 6 × 10¹4 Hz is shined on two slits separated a distance of 5 × 10-5 m. The detector
screen is located 1.5 m away.
a) Treating the slits as point-like, how far above the center of the detector is the first dark fringe? How about the first
bright fringe (excluding the bright fringe at the center)?
b) Consider the second dark fringe above the center of the detector and the first bright spot below the center of the
detector, how far apart are these locations?
c) Would your answer to a) increase, decrease, or stay the same if the light was violet? What if it was red?
d) For this part, suppose a small crystal is placed inside the top slit. The crystal generates a constant phase shift in
the light passing through that slit, but has no other effect. Now, at the center of the detector screen, there is a dark
fringe. How far above the center of the detector are the first and second bright fringes? How about the next dark fringe
(excluding the center)?
e) For this part, the crystal in part d) has been removed. Now let's consider the width of the slits themselves since, in
the real world, there is no such thing as a true point-like slit. Suppose the first dark fringe due to the single slit envelope
occurs at the same location as the third dark fringe due to the double slit interference pattern. What is the width of the
slits?
Transcribed Image Text:2. Green light with a frequency of 6 × 10¹4 Hz is shined on two slits separated a distance of 5 × 10-5 m. The detector screen is located 1.5 m away. a) Treating the slits as point-like, how far above the center of the detector is the first dark fringe? How about the first bright fringe (excluding the bright fringe at the center)? b) Consider the second dark fringe above the center of the detector and the first bright spot below the center of the detector, how far apart are these locations? c) Would your answer to a) increase, decrease, or stay the same if the light was violet? What if it was red? d) For this part, suppose a small crystal is placed inside the top slit. The crystal generates a constant phase shift in the light passing through that slit, but has no other effect. Now, at the center of the detector screen, there is a dark fringe. How far above the center of the detector are the first and second bright fringes? How about the next dark fringe (excluding the center)? e) For this part, the crystal in part d) has been removed. Now let's consider the width of the slits themselves since, in the real world, there is no such thing as a true point-like slit. Suppose the first dark fringe due to the single slit envelope occurs at the same location as the third dark fringe due to the double slit interference pattern. What is the width of the slits?
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“Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts for you. To get the remaining sub-parts solved please repost the complete question and mention the sub-parts to be solved.”

 

Interference of light waves through two slits: When the light wave from the same source is made to pass through two slits of point size separated by a distance d, then they form an interference pattern on the screen placed at D distance from the slits.

Position of mth maxima: The position of mth order maximum (or bright fringe) on the screen is given by the equation,

ym=mλDd                  . . . (1)

 where λ is the wavelength of light, y is the position of the bright fringe from the central maximum and m is an integer.

Position of mth minima: The position of mth order minimum (or dark fringe) on the screen is given by the equation,

ym=(2m+1)λD2d                            . . . (2)  

 where y is the position of the dark fringe from the central maximum and m is an integer.

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