I have a contradiction about refraction

[Efeguleroglu]

(Black one is the object and grey is its image.)
We know from Snell's Law:
$$
n_1\sin\alpha=n_2\sin\theta
$$
And I have been said that:
$$
a=b\ (1)\\\ and\\\ \frac{h}{h'}=\frac{n_2}{n_1} \ (2)
$$
Let's begin.
$$
\frac{n_1}{n_2}=\frac{\sin\alpha}{\sin\theta}=\frac{\sin(\frac{\pi}{2}-\alpha)}{\cos(\frac{\pi}{2}-\theta)}=\frac{b}{\sqrt{b^2+h'^2}} \frac{\sqrt{a^2+h^2}}{a}
$$
Since$$a=b$$
$$\frac{n_2}{n_1}=\frac{\sqrt{a^2+h^2}}{\sqrt{a^2+h'^2}}$$
And also $$\frac{n_2}{n_1}=\frac{h}{h'}$$
Thus $$\frac{h}{h'}=\frac{\sqrt{a^2+h^2}}{\sqrt{a^2+h'^2}}$$
But it's true only if $$h=h'$$
But that means $$n_1=n_2$$
And we're not observing this.
So which one is wrong? [(1) or (2)] (maybe not wrong but assumption)

 

[PeroK]

This is refraction. Not diffration.

How have you got $a = b$?

 

[Efeguleroglu]

Sorry corrected it.
It is always said so. That's why I am asking.

 

[PeroK]

Sorry corrected it.
It is always said so. That's why I am asking.

Assuming the two dots are supposed to be the same distance, $d$, from the impact point, then:

$\sin \alpha = \frac b d$

$\sin \theta = \frac a d$

 

[Efeguleroglu]

Then $$a\neq b$$ and $$\frac{h}{h'}=\frac{n_2}{n_1}=\frac{b}{a}$$

 

[Dale]

It is always said so.

Can you point to a specific source? I have never heard this, so I think that you must be misinterpreting some source.

 

[Efeguleroglu]

Can you point to a specific source? I have never heard this, so I think that you must be misinterpreting some source.

Must object and its image be on the same line that is perpendicular to the surface? And why? Actually that's what all I need to know. If yes (1) is correct (2) is wrong, if no (1) is wrong (2) is correct.

 

[Efeguleroglu]

Ok I got the problem.
$$a=b \\ \frac{h'}{h}=\frac{\tan\theta}{\tan\alpha}$$
That must be true. Thanks anyways.

 

[vanhees71]

[edited in view of #10; I'm always mixing up refraction (which is the case here) and diffraction ;-(]

I've no clue, how to interpret your drawing in #1. I think, it's on the refraction and reflection on a plane which is the boundary of two adjacent dielectrics with different indices of diffraction. For the reflection the reflected beam's direction is determined by the law reflection, according to which the reflected beam is in direction of the same angle with respect to the boundary's normal as the incoming beam, while for the refracted beam is directed according to Snell's law, which you quote correctly. I've no clue what the rest of your drawing means, particularly the gray and black points as well as the lengths $a$, $b$, $h$, and $h'$.

 

[Efeguleroglu]

I've no clue, how to interpret your drawing in #1. I think, it's on the diffraction and reflection on a plane which is the boundary of two adjacent dielectrics with different indices of diffraction. For the reflection the reflected beam's direction is determined by the law refrection, according to which the reflected beam is in direction of the same angle with respect to the boundary's normal as the incoming beam, while for the diffracted beam is directed according to Snell's law, which you quote correctly. I've no clue what the rest of your drawing means, particularly the gray and black points as well as the lengths $a$, $b$, $h$, and $h'$.

It is a refracted light ray. n1 and n2 are indexes. Black is object, grey is its image.

 

[vanhees71]

Yes, sorry its refraction, but I've still no clue what you mean by "object" and "image" in this context.

 

[PeroK]

Yes, sorry its refraction, but I've still no clue what you mean by "object" and "image" in this context.

The image is where the object appears to be to an observer in the top left of the diagram.

 

[Cutter Ketch]

a=b is true. To see where the image would appear you need more than one ray. The easiest second ray to consider is the ray normal to the surface. It does not refract. So the two rays of the image intersect on the normal ray from the object. Clearly a=b.

(2) is approximately true for small angles. That isn’t the problem.

You find that both statements can’t be true. The problem is in your analysis. Specifically the first thing you wrote is wrong. You wrote:

$\frac {n_1} {n_2} = \frac {sin \alpha} {sin \theta}$

That is wrong. You have inverted the relation. That is why you are getting a contradiction.

 

[sophiecentaur]

@Efeguleroglu your initial diagram, which contains an object and an image is not sufficient. It only tells you the directions in which a single ray will follow, To find the position of an image, you need two rays from the object which will produce a point at which the virtual image appears. If you search Google Images for 'ray diagrams', lenses, mirrors etc. you will see what I mean from the many hits. It is only in the most elementary treatments that you find the simplified sort of picture you have quoted and that isn't enough for you to get the main principle.