# Double slit experiment

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https://isaacphysics.org/questions/m...b-32642b4de0f3

Part B

The top ray travels root(102)/20m

The bottom one 0.5000089999m

^^^ using pythag

Difference of 4.966246999x10^-3m =1.25t-(0.5000089999-t)

t=0.224(433) but it says it’s wrong

Part B

The top ray travels root(102)/20m

The bottom one 0.5000089999m

^^^ using pythag

Difference of 4.966246999x10^-3m =1.25t-(0.5000089999-t)

t=0.224(433) but it says it’s wrong

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#2

(Original post by

https://isaacphysics.org/questions/m...b-32642b4de0f3

Part B

The top ray travels root(102)/20m

The bottom one 0.5000089999m

^^^ using pythag

Difference of 4.966246999x10^-3m =1.25t-(0.5000089999-t)

t=0.224(433) but it says it’s wrong

**Jsmithx**)https://isaacphysics.org/questions/m...b-32642b4de0f3

Part B

The top ray travels root(102)/20m

The bottom one 0.5000089999m

^^^ using pythag

Difference of 4.966246999x10^-3m =1.25t-(0.5000089999-t)

t=0.224(433) but it says it’s wrong

Is your t in m, cm or mm?

0.224(433)= 97 m, 97 cm or 9.7 mm?

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(Original post by

Try to explain the concepts that you are using instead of using numbers.

Is your t in m, cm or mm?

0.224(433)= 97 m, 97 cm or 9.7 mm?

**Eimmanuel**)Try to explain the concepts that you are using instead of using numbers.

Is your t in m, cm or mm?

0.224(433)= 97 m, 97 cm or 9.7 mm?

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#4

**Jsmithx**)

https://isaacphysics.org/questions/m...b-32642b4de0f3

Part B

The top ray travels root(102)/20m

The bottom one 0.5000089999m

^^^ using pythag

Difference of 4.966246999x10^-3m =1.25t-(0.5000089999-t)

t=0.224(433) but it says it’s wrong

(Original post by

Everything in m

**Jsmithx**)Everything in m

If you are marked correct, it would be ridiculous.

0.224(433)= 97 m is far larger than L = 50.0 cm

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#5

**Jsmithx**)

https://isaacphysics.org/questions/m...b-32642b4de0f3

Part B

The top ray travels root(102)/20m

The bottom one 0.5000089999m

^^^ using pythag

Difference of 4.966246999x10^-3m =1.25t-(0.5000089999-t)

t=0.224(433) but it says it’s wrong

*t*.

The number of waves in the glass of thickness

*t*of refractive index

*n*

_{g}:

Next, find the increase Δ

*N*in the number of waves in the glass when it replaces air of the same thickness,

*t*.

You are given how much the central band O is shifted down, and this is related to the product of Δ

*N*and fringe width.

Spoiler:

Δ

Show

Δ

*N ×*fringe width

*=*5.00 mm

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(Original post by

If you are marked correct, it would be ridiculous.

0.224(433)= 97 m is far larger than L = 50.0 cm

**Eimmanuel**)If you are marked correct, it would be ridiculous.

0.224(433)= 97 m is far larger than L = 50.0 cm

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(Original post by

Without the glass, the central band O is in the middle. The central band O is shifted towards Y by 5.00 mm in the presence of glass of thickness

The number of waves in the glass of thickness

Next, find the increase Δ

You are given how much the central band O is shifted down, and this is related to the product of Δ

**Eimmanuel**)Without the glass, the central band O is in the middle. The central band O is shifted towards Y by 5.00 mm in the presence of glass of thickness

*t*.The number of waves in the glass of thickness

*t*of refractive index*n*_{g}:Next, find the increase Δ

*N*in the number of waves in the glass when it replaces air of the same thickness,*t*.You are given how much the central band O is shifted down, and this is related to the product of Δ

*N*and fringe width.
Spoiler:

Δ

Show

Δ

*N ×*fringe width

*=*5.00 mm

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https://isaacphysics.org/questions/m...d-7e484c538940

16 micrometers was calculated by doing

λ = ax/D = nt/ΔN

ΔNx= 5x10^-3 m

a = 2x10^-3 m

D = 0.5 m

n= 1.25

What have I missed?

16 micrometers was calculated by doing

λ = ax/D = nt/ΔN

ΔNx= 5x10^-3 m

a = 2x10^-3 m

D = 0.5 m

n= 1.25

What have I missed?

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#10

**Eimmanuel**)

Without the glass, the central band O is in the middle. The central band O is shifted towards Y by 5.00 mm in the presence of glass of thickness

*t*.

The number of waves in the glass of thickness

*t*of refractive index

*n*

_{g}:

Next, find the increase Δ

*N*in the number of waves in the glass when it replaces air of the same thickness,

*t*.

You are given how much the central band O is shifted down, and this is related to the product of Δ

*N*and fringe width.

Spoiler:

Δ

Show

Δ

*N ×*fringe width

*=*5.00 mm

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(Original post by

n is the difference in refractive indexes of the 2 mediums so 0.25 not 1.25

**Sophhhowa**)n is the difference in refractive indexes of the 2 mediums so 0.25 not 1.25

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#12

(Original post by

Shouldn’t n(g) be n(medium)-n(air) so 0.25?

**Sophhhowa**)Shouldn’t n(g) be n(medium)-n(air) so 0.25?

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#13

(Original post by

I’ve equated this to ax/d then used delta N x = 5mm and got 16 micrometers as the ans. I put this into the site and it says ‘You need to consider the extra phase resulting from introducing the glass, when compared to what one would have if the glass were removed.’

**Jsmithx**)I’ve equated this to ax/d then used delta N x = 5mm and got 16 micrometers as the ans. I put this into the site and it says ‘You need to consider the extra phase resulting from introducing the glass, when compared to what one would have if the glass were removed.’

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#15

(Original post by

Refractive index of glass

**Sophhhowa**)Refractive index of glass

(Original post by

Shouldn’t n(g) be n(medium)-n(air) so 0.25?

**Sophhhowa**)Shouldn’t n(g) be n(medium)-n(air) so 0.25?

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#16

(Original post by

It make no sense for the refractive index of glass to be less than that of vacuum.

**Eimmanuel**)It make no sense for the refractive index of glass to be less than that of vacuum.

The refractive index of glass is still 1.25

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#17

(Original post by

I meant where you put ng in the formula it should have been n(glass)-n(air) so 0.25. By doing this you’ll get the correct answer.

The refractive index of glass is still 1.25

**Sophhhowa**)I meant where you put ng in the formula it should have been n(glass)-n(air) so 0.25. By doing this you’ll get the correct answer.

The refractive index of glass is still 1.25

What you are finding is the difference of the number of waves in glass and air NOT the number of waves in the glass (this is what the first maths expression implies).

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#18

(Original post by

https://isaacphysics.org/questions/m...d-7e484c538940

16 micrometers was calculated by doing

λ = ax/D = nt/ΔN

ΔNx= 5x10^-3 m

a = 2x10^-3 m

D = 0.5 m

n= 1.25

What have I missed?

**Jsmithx**)https://isaacphysics.org/questions/m...d-7e484c538940

16 micrometers was calculated by doing

λ = ax/D = nt/ΔN

ΔNx= 5x10^-3 m

a = 2x10^-3 m

D = 0.5 m

n= 1.25

What have I missed?

You are "blindly" applying the mathematical expressions or equation without really understanding the meaning of them.

is NOT the number of waves in the glass.

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#19

**Sophhhowa**)

n is the difference in refractive indexes of the 2 mediums so 0.25 not 1.25

**Sophhhowa**)

I meant where you put ng in the formula it should have been n(glass)-n(air) so 0.25. By doing this you’ll get the correct answer.

The refractive index of glass is still 1.25

What OP had done is (as far as I can see) putting everything together without understanding the physical meaning.

I did

**explicitly**state that

(Original post by

...Next, find the

**Eimmanuel**)...Next, find the

**increase Δ**when it replaces air of the same thickness,*N*in the number of waves in the glass*t*. ....
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