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Double slit experiment with a slit covered.

  • #1
304
14

Homework Statement


A thin flake of mica (n=1.58) is used to cover one slit of a double-slit arrangement.The central point on the screen is occupied by the 7th brigth fringe.If lamda = 550nm, what is the thickness of the mica?

Homework Equations


path difference d=(n-1)*thickness
d=7*lamda
snell's law?

The Attempt at a Solution


first, the paths above the glass must be the same.So they are not in consideration.
let l be the thickness of the glass,t be the length travelled by light in mica,x in air
The mica:
sinθ=nsin
sin∅=sinθ/n....(1)
sin∅=l/t..(2)

(1)/(2),
sinθ=nl/t...(3)

In the air:
sinθ=l/x...(4)
compare 4 and 3,
i came out, t=xn.

Then i started to consider the difference of wavelength,
because the wavelength in mica is shorter than in air.
I tried to subtract the sum of the pulses in air and mica and set it equals to 7 lamda,
but what came out is peculiar,
let m , n be the number of pulses in mica and air respectively,f=frequency
m=tf/v...(5)
n=xf/c..(6)
so....
m/n=nt/x
i suppose it being m/n = t/nx ......so that they can be cancelled out......
Please help for this:((
 

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
3,573
192

Homework Statement


A thin flake of mica (n=1.58) is used to cover one slit of a double-slit arrangement.The central point on the screen is occupied by the 7th brigth fringe.If lamda = 550nm, what is the thickness of the mica?

Homework Equations


path difference d=(n-1)*thickness
d=7*lamda
snell's law?

The Attempt at a Solution


first, the paths above the glass must be the same.So they are not in consideration.
let l be the thickness of the glass,t be the length travelled by light in mica,x in air
The mica:
sinθ=nsin
sin∅=sinθ/n....(1)
sin∅=l/t..(2)

(1)/(2),
sinθ=nl/t...(3)

In the air:
sinθ=l/x...(4)
compare 4 and 3,
i came out, t=xn.

Then i started to consider the difference of wavelength,
because the wavelength in mica is shorter than in air.
I tried to subtract the sum of the pulses in air and mica and set it equals to 7 lamda,
but what came out is peculiar,
let m , n be the number of pulses in mica and air respectively,f=frequency
m=tf/v...(5)
n=xf/c..(6)
so....
m/n=nt/x
i suppose it being m/n = t/nx ......so that they can be cancelled out......
Please help for this:((
I have the feeling that they do not expect you to take into consideration refraction through the mica (how could you, you do not have an angle of incidence). I think you can safely neglect that and just consider an incident angle of zero in which case what you wrote in part 2 is all you need.
 
  • #3
264
26
I don't quite like the way that this question is worded.
I think that it means that what was the 7th bright fringe becomes shifted into the central bright maximum.
Overall, I don't think that the appearance of the screen pattern would change.
Then the problem becomes to calculate the thickness of the mica in order for the
mica to contain 7 more wavelengths than would be contained in an equivalent thickness of air.
This should not be too difficult since you know that the wavelength in the mica = 5.5 * 10E-7 / 1.58.
as compared to 5.5 * 10E-7 in air.
 
  • #4
304
14
thank you all :)) i have solved the problem :D
 

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