# Geometric Optics- Refraction , reflected ray's

1. Jul 19, 2009

### sally143

1. The problem statement, all variables and given/known data

A beam of light in air makes an incident angle with the normal at 53 with an unknown substance. Part of the light is relected and part is refracted into the substance. The reflected ray and the refracted ray make an angle of 90 degrees.

a) What is the refracted angle?
b) what is the index of refraction for the substance?
c) the critical angle if the ray goes from the unknown substance into air?

2. Relevant equations
I think we use , Snells law nsin() = nsing () ; where () = Angle

3. The attempt at a solution

2. Jul 19, 2009

### tiny-tim

Welcome to PF!

Hi sally143! Welcome to PF!

Yes, that's ok so far …

now show us how far you get, and then we'll know how to help!

3. Jul 19, 2009

### sally143

Re: Welcome to PF!

Im confused, and not sure. But Incident angle is =53 degrees, refracted angle i think is =90degrees. i know that, , sin (theta).critical = n2/n1.

Water is 1.33
air is 1.0003

Not sure on where to start.

4. Jul 20, 2009

### Born2bwire

The angle of reflection is equal to the angle of incidence. The problem statement states that the angle between the reflected and refracted rays is 90 degrees. That should be enough to solve it from there.

5. Jul 20, 2009

### sally143

N2 and THeta 2, s (1.003)water sin 90?

ANd N.1 sing theta1, = N1 sin ().1 = (1)sin THeta2, solve for n1??

i need the steps im confused, trying to draw it out, but i get angle of incident
comng in, perpendicular . and angle of refraction on the right side. Both having
angles of 90 degrees.

Now to find criticle angle, its n2/n1 correct?

6. Jul 20, 2009

### Born2bwire

Why are you assuming water?

7. Jul 20, 2009

### sally143

I guess i assumed, because
i was up at 2:30am, and was
confused :), sorry

8. Jul 21, 2009

### Born2bwire

Yeah, the question seems to be asking you to figure out the angles of reflection and refraction and use these to find the index of refraction (parts (a) and (b)).

9. Jul 21, 2009

### sally143

Here the angle of incidence, i = 53 degree.whenever reflection takes place the angle of reflection will be equal to the angle of incidence.then the angle of reflection = 53 degree.
The angle between the reflected and the refracted ray = 90 degree
Therefore the angle between the refracted ray and the normal =the angle of refraction ,r = 180 - (53+90) = 180 -143 =37 degree.
A) = 37 degree.

B) Refractive index = sin i/sin r = sin 53/ sin 37 = 1.326
C) refractive index = 1/sin C ,where C is the critical angle or sin C =1/refractive index = 1/1.326 = 1/1.326 =0.7543
or C = sin inverse 0.7543 =48 degree 57 minutes.

does this make sense or this:

a)
Angle of refraction = 90° - 53° = 47°

b)
Refractive index
= sin(angle of incidence) / sin(angle of refraction)
= sin 53° / sin 47°
= 1.092.

c)
Critical angle
= sinֿ¹ (1 / 1.092)
= 66.3°.

10. Jul 22, 2009

### Born2bwire

Pay attention to how we reference the angles for incidence, reflection, and refraction in terms of the normals. The angles are always right angles or acute, as shown in the reflection and refration sections of Wikipedia's Gemetrical Optics page: http://en.wikipedia.org/wiki/Geometrical_optics . So draw a diagram of the problem, the angle between the refracted and reflected rays is 90 degrees. Use this to figure out the relationship of the refracted ray to calculated it's angle with the normal inside the second medium.