# Basic angles of refraction and incidence help needed

• Couperin
In summary, the basic angles of refraction and incidence help needed (help still needed). Someone needs to provide a visual representation of these concepts, as Quasar was not able to do so.
Couperin
Basic angles of refraction and incidence help needed (help still needed)

I feel silly for even asking this, but here goes.

http://img501.imageshack.us/img501/905/refractionpuzzlerg2.gif

The triangle represents a glass prism. The refractive index between air and the prism is 1.55. The angle of incidence (angle 'p' on my diagram) is 45 degrees, and by using the equation sin r = sin i / 1μ2, I've worked out angle 'q' to be roughly 21 degrees.

Knowing this, how could I work out angle 'r' on my diagram? I've tried visualising the two normal lines as two equals sides of an isoceles triangle, but I don't know whether that's just an illusion caused by the way I've drawn the diagram. I've also just assumed the angle 'r' is 90 degrees - 21 degrees, but I can't prove whether this is true or not.

So, could anyone assist me with working out angle 'r' on my diagram? A few hints or pointers?

Last edited by a moderator:
is the glass prism equilateral? If so, then the angle on the bottom left corner (let's call it A) is 60 ° and you can use the fact that the sum of the angles of the triangle Apr is 180°.

Thankyou, Quasar :).

So if I call that little triangle in the bottom left corner (ABC), and A = 60°, and B = 90° - q which = 69°, then angle C, which is next to r, must = 51°.

Therefore, the angle of incidence against the back surface of the prism is 90° - 51°, which is 39°.

So, the sin of the angle of refraction (the angle at which the light emerges from the prism) must be sin 39° / 1.55^-1, which = 0.98.

So the angle itself must be sin^-1 * 0.98, which = 77° (aprox).

So the initial incidence beam that hit the prism in the first place has, by entering the prism, diverged by...

...hmm, what's the proper way to calculate divergence? Do you add 77° to the first angle of refraction (21°)? If so, my answer is that it's diverged 98°, but I'm probably wrong.

## 1. What is the difference between refraction and incidence angles?

Refraction angle is the angle formed between the refracted ray and the normal line, while incidence angle is the angle formed between the incident ray and the normal line.

## 2. How are refraction and incidence angles related?

Refraction and incidence angles are related through Snell's Law, which states that the ratio of the sine of the incidence angle to the sine of the refraction angle is equal to the ratio of the speed of light in the first medium to the speed of light in the second medium.

## 3. What is the critical angle in refraction?

The critical angle in refraction is the angle of incidence where the refracted ray becomes parallel to the surface of the medium, resulting in total internal reflection.

## 4. How does the refractive index affect refraction angles?

The refractive index of a medium is a measure of how much the speed of light is reduced when it passes through that medium. A higher refractive index means that light will bend more when entering or exiting the medium, resulting in a larger refraction angle.

## 5. What are some real-life applications of refraction and incidence angles?

Refraction and incidence angles play important roles in various fields, such as optics, photography, and medicine. They are used in designing lenses, mirrors, and prisms, as well as in diagnosing eye conditions and capturing images in cameras and microscopes.

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