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The Physics PHYA2 thread! 5th June 2013

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What would the angle look like? Is it the same as when it hits the boundary? Also, where would the dotted line begin, exactly at the point at which the ray makes contact with the boundary? :smile:

I don't think you know the reflected angle exactly so an estimated guess (roughly the reflection angle should suffice).

Yes :smile:

(Correct me If I'm wrong guys)

Reply 1281

BigBadJFly

Original post by StalkeR47

But how do you know that if the incident substance has larger index than the other. And, incident angle is smaller than critical angle. TIR does not occur. How do you know this? Any explanations?

If the refractive index of n1 is less than n2, in the equation to work out the critical angle (sin thetac = n2/n1) , sin theta would be greater than 1, which is impossible

Reply 1282

StalkeR47

Original post by x-Sophie-x

Stalker, I'm not sure what you're trying to say exactly but I think you should accept that those guys are right haha :smile:

Sophie, I have accepted them. But, I am confused about the fact that, why TIR does not occur if the incident substance has larger refractive index than the emergent substance. I mean, you are only meeting 1 criteria (for TIR) but not the other. Why would it be refraction and not reflection. :smile:

Reply 1283

masryboy94

Original post by BigBadJFly

If the refractive index of n1 is less than n2, in the equation to work out the critical angle (sin thetac = n2/n1) , sin theta would be greater than 1, which is impossible

exactly

Reply 1284

kinglj81

please help me

Hi guys can you help me with something..

I was wondering what you do for questions about slits and diffraction that say show what you see on the screen. Do you draw the intensity graph type thing or just band or dots

Also when you diffract white light through a grating how do you show that the inner side is violet and the outer is red. Is this done on the intensity graph?.

Reply 1285

Qari

Original post by StalkeR47

Alright! Thanks man. Not really smart. I was just trying to clarify what I thought as in the exam paper question. And, I am still not sure. I know the rules properly but I need an explanations to why is that. :smile:

Thanks.

Explain to me why it is better for the signal to be in the core not the cladding?

Reply 1286

SpiritKing

Original post by StalkeR47

Amm sorry. I did not get you. Can you answer this please? Anyone please? why TIR does not occur if incident substance has larger refractive index than the other? Ca

otherwise sine of critical angle would exceed 1 which is not possible...

Reply 1287

masryboy94

Original post by StalkeR47

okay can i ask what you mean by incident substance, if you mean where the ray came from then what i have said earlier is correct.

Reply 1288

StalkeR47

Original post by BigBadJFly

If the refractive index of n1 is less than n2, in the equation to work out the critical angle (sin thetac = n2/n1) , sin theta would be greater than 1, which is impossible

Yes I know that. But in this case, n1 is not less than n2 is it?

Reply 1289

x-Sophie-x

Original post by StalkeR47

Because the angle of incidence doesn't exceed the critical angle? :confused:

I still don't understand the point you are trying to make entirely; I shall leave it to the experts :L

Reply 1290

Qari

Original post by StalkeR47

You've just stated it. You're only meeting 1 criteria :smile:

How do you work out critical angle? And what's the definition of critical angle?

(edited 10 years ago)

Reply 1291

x-Sophie-x

Original post by kinglj81

I'm not sure how, but I doubt you use intensity graphs to show that.

Reply 1292

StalkeR47

Original post by masryboy94

okay can i ask what you mean by incident substance, if you mean where the ray came from then what i have said earlier is correct.

Incident substance is the core.

Reply 1293

masryboy94

Original post by StalkeR47

No! All I was saying that why TIR does not occur if incident substance has larger refractive index than the other? Can you answer this? HUH?

Total internal reflection can happen only when a beam of light travelling through a dense medium crosses the interface with a rarer medium.
For example, through a glass piece to air.

When such a beam reaches an interface it makes an angle (called the angle of incidence) with the perpendicular at that point. When the beam exits the interface into the rarer medium. it makes a larger angle(called the angle of refraction) with the same perpendicular.

As the angle of incidence increases, so does the angle of refraction. There is one value of the angle of incidence for which the angle of refraction is 90 degrees and the emerging ray is tangential to the interface. This is called the critical angle.
For all angles of incidence greater than the critical angle the incident ray will not emerge into the rarer medium at all. In stead it gets reflected back into the denser medium itself.

Reply 1294

Qari

Original post by StalkeR47

Incident substance is the core.

How do you work out critical angle? And what's the definition of critical angle?

Reply 1295

Qari

Original post by x-Sophie-x

I'm not sure how, but I doubt you use intensity graphs to show that.

Do you have to show that? And wouldn't it be based on wavelengths?

Reply 1296

IWantSomeMushu

The grade boundaries on Unit 2 seem to be a lot lower than those of Unit 1.

Reply 1297

StalkeR47

Original post by Qari

You've just stated it. You're only meeting 1 criteria :smile:

How do you work out critical angle? And what's the definition of critical angle?

Alright then! sini(critical)=n2/n1. n2 is 1.45 and n1 is 1.55. critical angle is therefore 69.3 deg. So, 1 (criteria) is met and the other isn't. Then, why refraction and not reflection.?

Reply 1298

lebron_23

Original post by kinglj81

You would just show the screen how you would see it if you were looking at it. Meaning you would just draw a rectangle/square/your favourite shape and show the maxima as either the same width or half the width of the central maxima (depending on whether its double slit or single).

As for indicating the violet colour near to the central maxima, you would just label it saying something like 'the maxima closer to the central maxima have a violet tinge while the maxima further away take on an increasingly darker red tinge'.

Original post by IWantSomeMushu

The grade boundaries on Unit 2 seem to be a lot lower than those of Unit 1.

Yeah tell me about it! On average, you can drop 9 marks and still walk away with full UMS. Seems like a golden opportunity but I just know I'm going to throw away marks somewhere tomorrow