Refraction of light question -- Flintglass submerged in oil

[rssvn]

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?

 

[BvU]

Hi,

the question

What question ?

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

$\$

 

[rssvn]

Hi,

What question ?

$\$

oops, my bad, edited it to include the question :D

 

[BvU]

So the top $\alpha$ I can understand. But why and how the other one?

$\$

 

[Steve4Physics]

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.

 

[Steve4Physics]

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.

 

[rssvn]

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.

I don't quite understand why i need the b angle to answer the question

 

[Steve4Physics]

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?

 

[rssvn]

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?

 

[BvU]

I still don't see a complete problem statement ...

$\$

 

[Steve4Physics]

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..

 

[rssvn]

Yes..

Is the answer then (1.62sin(37))/1.4 -> arcsin(ans)

 

[Steve4Physics]

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.)

 

[rssvn]

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◦

 

[Steve4Physics]

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?

 

[rssvn]

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?

 

[Steve4Physics]

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.

 

[rssvn]

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?

 

[Steve4Physics]

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).

 

[rssvn]

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 :)