Presentation on theme: "CRITICAL ANGLE Natasha Solomon-Roy & Christina Iorio."— Presentation transcript:
CRITICAL ANGLE Natasha Solomon-Roy & Christina Iorio
PURPOSE To determine the critical angle of a semi-circular block that has an unknown index of refraction.
MATERIALS Lucite block (semi-circular) Ray Box (single slit) Polar co-ordinate paper Pencil
PROCEDURE 1. Lay the polar co-ordinated paper on a flat surface 2. Place the semi-circular block on the paper. In the center of the 0 o -180 o line. 3. Outline the block and draw a dot where the center is. 4. Darken the room and shine the ray to the dot previously made. 5. Move around the ray box until the ray can no longer refract and can only reflect. 6. Take down the angle 7. This will be the critical angle.
D OES NOT REFRACT IT REFLECTS ! Image 1
Diagram 1Diagram 2
FORMULAS Snell’s law: n 2. Sin c = n 1. Sin 90 o Sin c = n 1 n 2 Law 1: Law 2:
EXAMPLE Refractive indexes of medium A is 2, and medium B is 1,6. Find the critical angle of the rays coming from the medium A to B Diagram 3
DISSCUSSION Light rays travel through a highly refractive medium such as the semi-circular block and a weak refractive medium such as air. The light rays create total internal reflection of light which depends on the critical angle in relation to the index of refraction of the mediums. When shining the light at a 90 degree angle, total internal reflection occurs at the angle of incidence with respect to the normal.
ERRORS B lock moves out of place making it difficult to find the critical angle because it isn’t centered, therefore it is harder to have reflect the light. Light ray (from ray box) isn’t strong enough making it more difficult to see the reflected light.
C ONCLUSION What happened in this experiment? In this experiment, a highly refractive medium (semi-circular block) was examined with a single ray shining at the medium. The semi-circular block was to determine to reflect light when shone at 90 degrees because the critical angle created total internal reflection.
What did you learn from this experiment? Are there any applications for your findings? A highly refractive medium and a weak refractive medium, result in a critical angle which creates total internal reflection when at 90 degrees. The refracted ray stems across the interface when the angle of incidence is larger than the critical angle resulted in the reflection of light. This is relevant to fibre optics. In an optical fibre, such as the ones use in endoscopies, constantly reflect light at 90 degrees. The light waves of the optical fibre can travel at a greater distance because no light is absorbed which will result in a better view of patients examination.