Refraction of light question -- Flintglass submerged in oil

AI Thread Summary
The discussion revolves around calculating the angles of refraction for a laser beam passing through flintglass submerged in oil. Initial calculations using the refractive indices and angles led to confusion about the correct values and the need for a clear diagram. Participants emphasized the importance of accurately identifying angles and ensuring the diagram illustrates the light path effectively. Corrections were made to the calculations, leading to a confirmed angle of 44 degrees for θ₁. The conversation concluded with a focus on improving the diagram for clarity and accuracy in representing the refraction process.
rssvn
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Homework Statement
A container hold oil with refraction index of 1.4. A flintglass? (maybe a wrong translation) is submerged into the oil, with a incline of a=37◦ flintglass index is 1.62. Laser is shot at the flintglass which then exits into the oil, what is the angle when the laser exits the glass?
Relevant Equations
n1sinΘ1=n2sinΘ2
Hello, hopefully the question made sense, it was hard to translate. i attached a photo about the question.

I started with n1=1.4, sinΘ1=37◦ and n2=1.62

1.4(sin(37◦))=1.62sinΘ2
1.4(sin(37◦))/1.62=sinΘ2
arcsin(0.52)=31.34◦

Is it calculated correctly?
 

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Hi,
rssvn said:
the question
What question ?

Also: I see ##\alpha, \ \beta,\ ## but no ##\theta_1,\ \theta_2\ ## ... ?

##\ ##
 
BvU said:
Hi,

What question ?

##\ ##
oops, my bad, edited it to include the question :D
 
So the top ##\alpha## I can understand. But why and how the other one?

##\ ##
 
If you can find ##\beta##, you have nearly completed the problem!

It is not clear (as noted by @BvU ) what angles ##\theta_1## and ##\theta_2## are.
 
BvU said:
So the top ##\alpha## I can understand. But why and how the other one?
Hi @BvU. I'm guessing that the top of the block is inclined clockwise by ##\alpha## to the vertical and the ray inside the block is traveling horizontally. As a result the ray inside the block makes an angle ##\alpha## to the normal. Presumably it's just the way the person who wrote the question decided to set it up.
 
BvU said:
So the top ##\alpha## I can understand. But why and how the other one?

##\ ##
Steve4Physics said:
Hi @BvU. I'm guessing that the top of the block is inclined clockwise by ##\alpha## to the vertical and the ray inside the block is traveling horizontally. As a result the ray inside the block makes an angle ##\alpha## to the normal. Presumably it's just the way the person who wrote the question decided to set it up.
I don't quite understand why i need the b angle to answer the question
 
rssvn said:
I don't quite understand why i need the b angle to answer the question
You should look at this diagram very carefully: https://www.gcsescience.com/light-refraction-glass.gif (Having oil rather than air doesn't matter here.)

Can you understand why the emergent ray is parallel to the incident ray?

Does this help you answer the question?
 
Steve4Physics said:
You should look at this diagram very carefully: https://www.gcsescience.com/light-refraction-glass.gif (Having oil rather than air doesn't matter here.)

Can you understand why the emergent ray is parallel to the incident ray?

Does this help you answer the question?
Ooh, so because the flintglass is submerged, the laser must have been traveling in the oil already before it hit the flintglass, and since the b angle is angle it hits the flintglass it must come out in the same angle too?
 
  • #10
I still don't see a complete problem statement ...

##\ ##
 
  • #11
rssvn said:
Ooh, so because the flintglass is submerged, the laser must have been traveling in the oil already before it hit the flintglass, and since the b angle is angle it hits the flintglass it must come out in the same angle too?
Yes..
 
  • #12
Steve4Physics said:
Yes..
Is the answer then (1.62sin(37))/1.4 -> arcsin(ans)
 
  • #13
rssvn said:
Is the answer then (1.62sin(37))/1.4 -> arcsin(ans)
I can't say. (Edit: note 'arcsin(ans)' makes no sense.)

Which angle of the laser beam are you calculating? The angle to the normal at the exit point? Or the angle to the horizontal? Or the angle to to the vertical? Other something else?
(You should have a diagram clearly showing the full path of the laser beam and the angles.)

If you start with the ‘standard equation’:
n₁sin θ₁ = n₂sinθ₂
what values are you using for each variable? Can you clearly show all the working, including the final answer?

Here are some symbols you can cut and paste to make your reply easier to follow: n₁ θ₁ n₂ θ₂ α β. (I haven’t used Latex which is the usual method here.)
 
  • #14
Steve4Physics said:
I can't say. (Edit: note 'arcsin(ans)' makes no sense.)

Which angle of the laser beam are you calculating? The angle to the normal at the exit point? Or the angle to the horizontal? Or the angle to to the vertical? Other something else?
(You should have a diagram clearly showing the full path of the laser beam and the angles.)

If you start with the ‘standard equation’:
n₁sin θ₁ = n₂sinθ₂
what values are you using for each variable? Can you clearly show all the working, including the final answer?

Here are some symbols you can cut and paste to make your reply easier to follow: n₁ θ₁ n₂ θ₂ α β. (I haven’t used Latex which is the usual method here.)
I'm trying to solve the β angle,
n₁=1.4,
n₂=1.62,
θ₂=37◦,
θ₁=?
n₁sinθ₁ = n₂sinθ₂
sinθ₁=(n₂sinθ₂)/n₁
θ₁=sin-1((n₂sinθ₂)/n₁)

θ₁=sin-1((1.4sin(37◦))/1.62)

θ₁=44.14◦
 
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  • #15
rssvn said:
I'm trying to solve the β angle,
n₁=1.4,θ₁=sin-1((n₂sinθ₂)/n₁)
n₂=1.62,
θ₂=37◦,
θ₁=?
n₁sinθ₁ = n₂sinθ₂
sinθ₁=(n₂sinθ₂)/n₁
θ₁=sin-1((n₂sinθ₂)/n₁)

θ₁=sin-1((1.4sin(37◦))/1.62)

θ₁=44.14◦
Good try. But 4 problems!

1) Your equation θ₁=sin⁻¹((n₂sinθ₂)/n₁) does not match θ₁=sin⁻¹((1.4sin(37◦))/1.62). Check your values.

2) sin⁻¹((1.4sin(37◦))/1.62) works out to be 31º, not 44º. (But 31º is wrong!)

(When you do it correctly, you will find mistakes 1 and 2 cancel!)

3) You have the wrong number of significant figures in your anwer.

4) If this is a written assignment to be handed in, you must include a diagram. Make it clear what angle θ₁ is on your diagram, or your teacher can not tell if you have found the correct angle to answer the original question. Can you post your diagram here?
 
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  • #16
Steve4Physics said:
Good try. But 4 problems!

1) Your equation θ₁=sin⁻¹((n₂sinθ₂)/n₁) does not match θ₁=sin⁻¹((1.4sin(37◦))/1.62). Check your values.

2) sin⁻¹((1.4sin(37◦))/1.62) works out to be 31º, not 44º. (But 31º is wrong!)

(When you do it correctly, you will find mistakes 1 and 2 cancel!)

3) You have the wrong number of significant figures in your anwer.

4) If this is a written assignment to be handed in, you must include a diagram. Make it clear what angle θ₁ is on your diagram, or your teacher can not tell if you have found the correct angle to answer the original question. Can you post your diagram here?
Oh seems that i forgot to insert the right values, it was meant to be: θ₁=sin-1((1.62sin(37◦))/1.4)
and the answer for β would be 44º (i think significant figures mean the answers accuracy? so to say)
Would the diagram in the attached photo be sufficient?
 

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  • #17
rssvn said:
Oh seems that i forgot to insert the right values, it was meant to be: θ₁=sin-1((1.62sin(37◦))/1.4)
and the answer for β would be 44º (i think significant figures mean the answers accuracy? so to say)
Would the diagram in the attached photo be sufficient?
θ₁=sin⁻¹((1.62sin(37◦))/1.4) - good
44º - good
Diagram - bad!

Going from left-to-right:
Laser beam passes through oil and is refracted at oil/glass boundary.
Laer beam passes through glass.
What does the laser beam do next? You must add this on the diagram. Include arrows on the diagram to show which direction the beam is travelling. Show the angle you are being asked to find.
 
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  • #18
Steve4Physics said:
θ₁=sin⁻¹((1.62sin(37◦))/1.4) - good
44º - good
Diagram - bad!

Going from left-to-right:
Laser beam passes through oil and is refracted at oil/glass boundary.
Laer beam passes through glass.
What does the laser beam do next? You must add this on the diagram. Include arrows on the diagram to show which direction the beam is travelling. Show the angle you are being asked to find.
Would this be better?
 

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  • #19
rssvn said:
Would this be better?
Yes – good!

The only improvements I can suggest are:
- draw arrow-heads on the light rays, as shown here: https://www.gcsescience.com/light-refraction-glass.gif
- take a bit more care drawing the normal accurately; a normal should be 90º to the surface (yours is somewhere between 80º and 90º).
 
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  • #20
Steve4Physics said:
Yes – good!

The only improvements I can suggest are:
- draw arrow-heads on the light rays, as shown here: https://www.gcsescience.com/light-refraction-glass.gif
- take a bit more care drawing the normal accurately; a normal should be 90º to the surface (yours is somewhere between 80º and 90º).
Good tips, i will definitely remember them. Thank you for all the help, I've learned a lot :)
 
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