# Homework Help: Young's double slit experiment beams

1. Mar 31, 2009

### rayhan619

1. The problem statement, all variables and given/known data
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

2. Relevant 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 im doing wrong. and how do I get m?

3. The attempt at a solution

2. Apr 1, 2009

### Redbelly98

Staff Emeritus
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. Apr 1, 2009

### rayhan619

I have attached the pic here. you can take a look.
thank you

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4. Apr 2, 2009

### Redbelly98

Staff Emeritus
Thanks.

My comment in post #2 applies here. What is the wavelength within the glass?

5. Apr 2, 2009

### rl.bhat

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. Apr 2, 2009

### rayhan619

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 whats theta in here?
its not given

7. Apr 2, 2009

### rl.bhat

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. Apr 2, 2009

### rayhan619

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. Apr 2, 2009

### Redbelly98

Staff Emeritus
If there are 2 bright fringes on either side of the central (θ=0) fringe, how many does that add up to?

10. Apr 2, 2009

### rayhan619

so the total m is 5.
right?
how do i do part b) ?

11. Apr 2, 2009

### Redbelly98

Staff Emeritus
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. Apr 2, 2009

### rayhan619

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. Apr 2, 2009

### Redbelly98

Staff Emeritus
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.

Yes.

14. Apr 2, 2009

### rl.bhat

First of all find the critical angle of the glass. Then decide which ray coming out of the glass.

15. Apr 2, 2009

### rayhan619

so the critical angel is = sin^-1(n2/n1)
= sin^-1(1.00/1.52)
= 41.14 degree
that is the answer.right?

16. Apr 2, 2009

### rayhan619

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

whats the next step?

17. Apr 2, 2009

### rl.bhat

You have calculated two angles. Out of which only one comes out of the glass. Which one?

18. Apr 2, 2009

### rayhan619

so only the first angel comes out because its less than 41 degree.

19. Apr 2, 2009

### rl.bhat

Yes. In all three fringes come out.

20. Apr 2, 2009

### rayhan619

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?