Double slit problem with glass block

AI Thread Summary
Light of wavelength 600 nm passing through two slits separated by 0.20 mm creates an interference pattern observed on a screen 1.0 m away. The delay caused by a thin piece of glass in one slit is 5.0x10^-16 seconds, which needs to be converted into a fraction of the light wave's period to determine the phase difference. The phase difference can be calculated based on the delay and the wavelength, affecting the interference pattern. The introduction of glass, which has a higher refractive index than air, is expected to shift the central maximum away from the slit containing the glass. Further calculations are needed to quantify this shift accurately.
Sarah Hallsway
Messages
3
Reaction score
0

Homework Statement


Light of wavelength 600 nm passes through two slits separated by 0.20 mm and is observed on a screen 1.0 m behind the slits. The location of the central maximum is marked on the screen and labeled y=0.

(I only need help on the last 2 parts, but I will list all of them in case information from them is needed to complete the last parts.)

a. At what distance, on either side of y=0, are the m=1 bright fringes?

b. A very thin piece of glass is then placed in one slit. Because light travels slower in glass than in air, the wave passing through the glass is delayed by 5.0x10^-16 s in comparison to the wave going through the other slit.
What fraction of the period of the light wave is this delay?

c. With the glass in place, what is the phase difference, [delta phi][/0], between the two waves as they leave the slits?

d. The glass causes the interference fringe pattern on the screen to shift sideways. Which way does the central maximum move (toward or away from the slit with the glass) and by how far?

Homework Equations


lambda=yd/mL
where
y= place on y-axis of screen
d= distance between two slits
m=diffraction order
L= distance of slits from screen

and
period=1/frequency

lambda=v/f

The Attempt at a Solution


For the first problem, I plugged in the known values into the first equation I provided, and got an answer of 3.0 mm. For the second problem, I manipulated some variables in the bottom two equations to get a fraction of a period.

I am not stuck on the last two. For c, I started by dividing 1.0m/600E-9m to get how many wavelengths ahead one light beam is from the other. That value was 1.6E6. I took this to mean that without the glass, this value represented how far ahead the wave is from the other. However, without being given the thickness of the glass, I do not know how to take it into account.

For d, I am having a similar problem, since the thickness of the glass is unknown. Logically, I would imagine that the fringes would move away from the slit with the glass, since the glass has a higher n value than air does. To solve this problem, I drew a triangle from the two slits to the fringes, but could not find enough known values to calculate y.

Any help is appreciated. Thank you!
 
Physics news on Phys.org
Sarah Hallsway said:
For c, I started by dividing 1.0m/600E-9m
You are asked about the phase difference caused by the glass. Why would the distance from there to the screen be relevant?
 
What is the distance the "unencumbered" wave has traveled from the slit when
the wave that was delayed by 5 * 10 E-16 sec emerges from its slit?
What fraction of a wavelength is this?
How does this affect the geometry at the slits?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top