Refraction of light question -- Flintglass submerged in oil

In summary, the equation θ₁=sin⁻¹((n₂sinθ₂)/n₁) does not match θ₁=sin⁻¹((1.4sin(37◦))/1.62). Check your values.
  • #1
rssvn
26
4
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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  • #2
Hi,
rssvn said:
the question
What question ?

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

##\ ##
 
  • #3
BvU said:
Hi,

What question ?

##\ ##
oops, my bad, edited it to include the question :D
 
  • #4
So the top ##\alpha## I can understand. But why and how the other one?

##\ ##
 
  • #5
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.
 
  • #6
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.
 
  • #7
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
 
  • #8
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?
 
  • #9
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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Related to Refraction of light question -- Flintglass submerged in oil

What is refraction of light?

Refraction of light is the bending of light as it passes through a medium with a different optical density. This is due to the change in speed of light as it travels from one medium to another.

How does refraction occur in a Flintglass submerged in oil?

When light passes from air into a denser medium, such as oil, it slows down and bends towards the normal. When it passes from oil into a less dense medium, such as Flintglass, it speeds up and bends away from the normal. This results in a change in the direction of the light ray, causing refraction.

What is the role of the angle of incidence in refraction?

The angle of incidence is the angle at which the light ray hits the surface of the medium. It determines the amount of bending that will occur in the light ray as it passes through the medium. The larger the angle of incidence, the greater the amount of bending.

Why does the speed of light change in different media?

The speed of light changes because the particles in the medium interact with the light waves, causing them to slow down or speed up. This change in speed is what causes the light to bend as it passes through different media.

How does the refractive index of a medium affect the amount of refraction?

The refractive index is a measure of how much a medium can slow down light. The higher the refractive index, the slower the speed of light in that medium and the greater the amount of refraction. This is why different media have different amounts of refraction.

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