Young's double slit experiment beams

In summary: Put these values in the formula and find theta. So theta when m=1 is sin^-1(1*434.2 nm/1*10^-6 m) = 25.7 degreeand when m= 2, theta is sin^-1(2*434.2 nm/1*10^-6 m) = 60.3 degree
  • #1
rayhan619
77
0

Homework Statement


In a Young's double slit experiment, a pair of slits is encased in a rectangular block of glass (n=1.52), and glass block is surrounded by air, as shown below. The glass block is illuminated by coherent light (k = 660 nm) from a laser, as shown. The beam enters the block at normal incidence, propagates through the glass and strikes the slit-pair, which are separated by W = lxlO"6m. The wavelets emerging from the slit undergo interference, as usual, (a) At most, how many bright fringes can be formed on eitherside of the bright fringe, and what are the angles associated with these bright fringes? Sketch the situation, (b) How many primary beams emerge from the glass block on the side of the observer, and what are the angles at which they emerge?
sin theta = lamda/d

theta = lamda/d

Homework Equations



sin theta = lamda/d
theta = sin^-1 (lamda/d)
= sin^-1 (660*10^-9/1*10^-6)
= 41.29 degree

Theta = lamda/d
= (660*10^-9/1*10^-6)
= 0.66 degree

Im not sure what I am doing wrong. and how do I get m?



The Attempt at a Solution

 
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  • #2
I can't see your figure, but if the diffraction occurs within the glass then you need to use the wavelength within the glass, which is different than the 660 nm wavelength in air.
 
  • #3
I have attached the pic here. you can take a look.
thank you
 

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  • #4
Thanks.

My comment in post #2 applies here. What is the wavelength within the glass?
 
  • #5
Refractive index n = Lambda( in air)/lambda( in glass)
Find lambda in glass.
For bright fringes sin(theta) = m*lambda/d <1.
Find m. From that find theta.
 
  • #6
n = lamda air/ lamda glass
lamda glass = lamda air/ n
= 660/1.52
= 434.2

for bright fringes, sin theta = m*lamda/d
m = sin theta*d/lamda

but what's theta in here?
its not given
 
  • #7
First of all find the maximum value of m. Sin(theta) cannot be greater than one.
So find the value of d/lambda. = 1x10-6/434.2 nm = 2.3.
Hence m can be 1 or 2. Put these values in the formula and find theta.
 
  • #8
so the theta when m=1 is
sin^-1(1*434.2 nm/1*10^-6 m) = 25.7 degree
and when m= 2, theta is
sin^-1(2*434.2 nm/1*10^-6 m) = 60.3 degree

so the ans of part a) is 2 bright fringes can form on either side with angels of 25.7 and 60.3 degrees.

Right?

then how do I do part b) ?
 
  • #9
If there are 2 bright fringes on either side of the central (θ=0) fringe, how many does that add up to?
 
  • #10
so the total m is 5.
right?
how do i do part b) ?
 
  • #11
You now have part of part b, "How many primary beams emerge from the glass block on the side of the observer".

You also have the angles of the beam fringes within the glass. You just need to figure out what those angles become when those beams emerge from the glass into the air.
 
  • #12
so for both part a) and b) the number of beam are same which is 5?
can we get the angels for part b using
n(air)*sintheta(air) = n(glass)*Sintheta(glass) ?
 
  • #13
rayhan619 said:
so for both part a) and b) the number of beam are same which is 5?
Not quite. There are 5 beams in total, which is what (b) asks for.
In (a), they ask how many beams are on either side of the central bright fringe. And as you saw, there are 2.

can we get the angels for part b using
n(air)*sintheta(air) = n(glass)*Sintheta(glass) ?
Yes.
 
  • #14
First of all find the critical angle of the glass. Then decide which ray coming out of the glass.
 
  • #15
so the critical angel is = sin^-1(n2/n1)
= sin^-1(1.00/1.52)
= 41.14 degree
that is the answer.right?
 
  • #16
so the critical angel is
=sin^-1(n2/n1)
=sin^-1(1.00/1.52)
= 41.14 degree

whats the next step?
 
  • #17
rayhan619 said:
so the theta when m=1 is
sin^-1(1*434.2 nm/1*10^-6 m) = 25.7 degree
and when m= 2, theta is
sin^-1(2*434.2 nm/1*10^-6 m) = 60.3 degree

so the ans of part a) is 2 bright fringes can form on either side with angels of 25.7 and 60.3 degrees.

Right?

then how do I do part b) ?
You have calculated two angles. Out of which only one comes out of the glass. Which one?
 
  • #18
so only the first angel comes out because its less than 41 degree.
 
  • #19
Yes. In all three fringes come out.
 
  • #20
n = lamda air/ lamda glass
lamda glass = lamda air/ n
= 660/1.52
= 434.2

for bright fringes, sin theta = m*lamda/d
m = sin theta*d/lamda
d/lambda. = 1x10-6/434.2 nm = 2.3. [Sin(theta)<1]
m can be 1 or 2

so the theta when m=1 is
sin^-1(1*434.2 nm/1*10^-6 m) = 25.7 degree
and when m= 2, theta is
sin^-1(2*434.2 nm/1*10^-6 m) = 60.3 degree

the critical angel is = sin^-1(n2/n1)
= sin^-1(1.00/1.52)
= 41.14 degree

only the first angel comes out because its less than 41 degree.
And so total 3 fringes come out of the glass.

Is that the whole solution?
so for both part a) and b) it is 3 fringes?
 
  • #21
so for both part a) and b) it is 3 fringes?
No. For a part 5 fringes.
 

1. What is the Young's double slit experiment?

The Young's double slit experiment is a famous physics experiment that demonstrates the wave-like nature of light. It involves shining a beam of light through two parallel slits and observing the interference pattern that is created on a screen behind the slits.

2. How does the double slit experiment demonstrate the wave nature of light?

The interference pattern created in the double slit experiment is caused by the wave-like behavior of light. When light passes through the two slits, it diffracts and creates overlapping waves, resulting in the pattern of bright and dark fringes on the screen.

3. What is the significance of the interference pattern in the double slit experiment?

The interference pattern in the double slit experiment provides evidence for the wave-particle duality of light. It shows that light behaves like a wave, but can also act as a particle, depending on how it is observed.

4. Can the double slit experiment be performed with other types of particles besides light?

Yes, the double slit experiment has been performed with other types of particles such as electrons, protons, and even large molecules like buckyballs. This further supports the wave-particle duality concept in quantum mechanics.

5. What practical applications does the double slit experiment have?

The double slit experiment has practical applications in fields like optics, where it is used to study diffraction and interference phenomena. It is also a fundamental concept in quantum mechanics and has been used to study the behavior of particles at the atomic and subatomic level.

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