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Lesson 5. Problem: Light enters a prism as shown, and passes through the prism. a)Complete the path of the light through the prism, and show the angle.

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Presentation on theme: "Lesson 5. Problem: Light enters a prism as shown, and passes through the prism. a)Complete the path of the light through the prism, and show the angle."— Presentation transcript:

1 Lesson 5

2 Problem: Light enters a prism as shown, and passes through the prism. a)Complete the path of the light through the prism, and show the angle it will make when it leaves the prism. b)If the refractive index of the glass is 1.55, calculate the angle of refraction when it leaves the prism. c)How would the answer to b) change if the prism were immersed in water? 30 o 60 o glass air

3 Solution: Light enters a prism as shown, and passes through the prism. a)Complete the path of the light through the prism, and show the angle it will make when it leaves the prism. b)If the refractive index of the glass is 1.55, calculate the angle of refraction when it leaves the prism. c)How would the answer to b) change if the prism were immersed in water? 30 o 60 o glass air 30 o 22 c) In water, the angle of the light as it leaves the glass would be smaller, since the indices of refraction would be more similar and there would be less bending.

4 Problem: Light enters a prism made of air from glass. a)Complete the path of the light through the prism, and show the angle it will make when it leaves the prism. b)If the refractive index of the glass is 1.55, calculate the angle of refraction when it leaves the prism. 30 o 60 o glass air

5 Problem: Light enters a prism made of air from glass. a)Complete the path of the light through the prism, and show the angle it will make when it leaves the prism. b)If the refractive index of the glass is 1.55, calculate the angle of refraction when it leaves the prism. 60 o glass air 30 o 22

6  The smallest angle of incidence for which light cannot leave a medium is called the critical angle of incidence.  If light passes into a medium with a lesser refractive index than the original medium, it bends away from the normal and the angle of refraction is greater than the angle of incidence.  If the angle of refraction is > 90 o, the light cannot leave the medium.

7 This drawing reminds us that when light refracts from a medium with a larger n into one with a smaller n, it bends away from the normal. n1n1 n2n2 n 1 > n 2

8 This shows light hitting a boundary at the critical angle of incidence, where the angle of refraction is 90 o. No refraction occurs! n1n1 n2n2 cc  r = 90 o n 1 > n 2

9 Instead of refraction, total internal reflection occurs when the angle of incidence exceeds the critical angle. n1n1 n2n2 cc  r = 90 o n 1 > n 2 Ray reflects instead of refracts.

10  Simple use Snell’s Law to calculate the critical angle of incidence.  The angle in medium 2 is set to 90 o, and  1 is the critical angle of incidence.  n 1 sin  1 =n 2 sin(90 o ).

11 Binoculars use a combination of prisms that reflect the incoming light. As long as the incident angles exceed the critical angle, the light will be reflected.

12 Light enters the fiber optic tube at an angle above the critical angle and is thus totally reflected down the ‘light pipe’ to the other end. For commercial use, two different glasses are used, wrapped in a protective cover. Which must have the greater index of refraction, the core or the cladding?

13 Problem: What is the critical angle of incidence for a gemstone with refractive index 2.45 if it is in air? If you immerse the gemstone in water (refractive index 1.33), what does this do to the critical angle of incidence?

14 Solution: What is the critical angle of incidence for a gemstone with refractive index 2.45 if it is in air? If you immerse the gemstone in water (refractive index 1.33), what does this do to the critical angle of incidence? It increases the critical angle of incidence because there is less difference in the refractive indices.

15 Solution: The glass core of an optical fiber has an index of refraction of 1.60. The index of refraction of the cladding is 1.48. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?

16 Problem: The glass core of an optical fiber has an index of refraction of 1.60. The index of refraction of the cladding is 1.48. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?

17  Different colors of light have slightly different refractive indices.  Therefore, different colors composing white light will bend at slightly different angles when refracting through glass.  This causes the “prism” effect. physics.ohio-state.edu


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