Jsmithx
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#1
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#1
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
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Eimmanuel
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(Original post by 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
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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Jsmithx
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(Original post by 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?
Everything in m
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Eimmanuel
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#4
(Original post by 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 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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Eimmanuel
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(Original post by 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
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 ng:


 \dfrac{t n_g}{\lambda }

Next, find the increase ΔN 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 ΔN and fringe width.

Spoiler:
Show

ΔN × fringe width = 5.00 mm
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Jsmithx
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#6
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#6
(Original post by Eimmanuel)
If you are marked correct, it would be ridiculous.
0.224(433)= 97 m is far larger than L = 50.0 cm
The 433 are the following insignificant figures. 97m is not my ans 0.224m is
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Jsmithx
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(Original post by 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 ng:


 \dfrac{t n_g}{\lambda }

Next, find the increase ΔN 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 ΔN and fringe width.

Spoiler:
Show

ΔN × fringe width = 5.00 mm
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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Jsmithx
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#8
https://isaacphysics.org/questions/m...d-7e484c538940

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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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Sophhhowa
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n is the difference in refractive indexes of the 2 mediums so 0.25 not 1.25
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Sophhhowa
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#10
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(Original post by 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 ng:


 \dfrac{t n_g}{\lambda }

Next, find the increase ΔN 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 ΔN and fringe width.

Spoiler:
Show

ΔN × fringe width = 5.00 mm
Shouldn’t n(g) be n(medium)-n(air) so 0.25?
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Jsmithx
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(Original post by Sophhhowa)
n is the difference in refractive indexes of the 2 mediums so 0.25 not 1.25
Thank you!
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Eimmanuel
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(Original post by Sophhhowa)
Shouldn’t n(g) be n(medium)-n(air) so 0.25?
What is n(g)?
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Eimmanuel
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(Original post by 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.’
The first mathematical expression in post #5 is NOT delta N.
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Sophhhowa
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(Original post by Eimmanuel)
What is n(g)?
Refractive index of glass
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Eimmanuel
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(Original post by Sophhhowa)
Refractive index of glass
(Original post by Sophhhowa)
Shouldn’t n(g) be n(medium)-n(air) so 0.25?
It make no sense for the refractive index of glass to be less than that of vacuum.
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Sophhhowa
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(Original post by Eimmanuel)
It make no sense for the refractive index of glass to be less than that of vacuum.
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
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Eimmanuel
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(Original post by 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
By getting the correct answer does not imply the first maths expression in the post #5 is incorrect. You need to read carefully what it means.

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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Eimmanuel
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(Original post by Jsmithx)
https://isaacphysics.org/questions/m...d-7e484c538940

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Views: 23
Size:  262.2 KB

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?
Don't post the same problem twice.

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

 \Delta N is NOT the number of waves in the glass.
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Eimmanuel
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(Original post by Sophhhowa)
n is the difference in refractive indexes of the 2 mediums so 0.25 not 1.25
(Original post by 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
Your symbol n is different from what I state in the maths expression in post #5.

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 Eimmanuel)
...Next, find the increase ΔN in the number of waves in the glass when it replaces air of the same thickness, t. ....
Similar symbols have different meaning and can be confusing. The meaning should be derived from the context of writing.
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