This document discusses keyhole surgery and fiber optics. It provides two key points: 1) Keyhole surgery involves using two optical fibers - one to illuminate the inside of the patient and one for a camera to send images back to the doctor, allowing surgery to be performed through small incisions. 2) In medieval times, surgeons would open up patients to see what was wrong, whereas modern keyhole surgery uses these optical fibers to perform stomach cancer operations through small incisions.

1. Refraction and total internal reflection Lesson 2

2. The Endoscope There are two optical fibres One for light, to illuminate the inside of the patient One for a camera to send the images back to the doctor. Key hole surgery

3. Step into the photo You are the patient in the photos Describe your operation in medieval times Describe you operation in modern times with keyhole surgery Surgeons at Worthing and Southlands pioneer ‘keyhole’ operating techniques for the treatment of stomach cancers. Medieval surgeons open up a patient To see what is wrong.

4. Fibre optics

5. n = 2.41 Diamonds sparkle Calculate the critical angle for a diamond Sin θ c = n 2 /n 1 θ c = 24.5

6. Snell’s Law The refractive index (n) and v explained on next slide

7. What is the refractive index The refractive index (n)of a medium is the speed of light in vacuum (C) divided by the speed of light in the medium (V) It is the "optical density" of the medium. It is always greater than 1 n = C/V Vacuum: n = 1 exactly Air: n = 1.0003, often approximated by 1 Water: n = 1.34, average over visible range Glass: n = 1.5 is typically used in optics homework problems Fused silica (pure SiO2 glass): n = 1.4 Common refractive indexes

8. Quick Refraction The speed of light in a certain glass is 1.8  10 8 m s -1 . What is the refractive index of the glass? The refractive index of diamond is 2.4. What is the speed of light in diamond? 3. The refractive index changes with the colour of the light leading to dispersion. If the refractive index for blue light in a certain glass is 1.639 and for red light is 1.621, calculate the angle between the rays if they were both incident at 50 o n = C/V C = 3 x 10 8 in air n (air) = 1

9. Correct Answers n g = c/v g = 3.0  10 8 / 1.8  10 8 = 1.67 (no unit as refractive index is a number) n d = c /v d so v d = c/n d = 3  10 8 m s -1 / 2.4 = 1.25  10 8 m s -1 1 sin θ 1 = n 2 sin θ 2 3. For blue light, sin θ 1 = sin 50° / 1.639 = 0.4674 and θ = 27.86 o For red light, sin θ 1 = sin 50° / 1.621 = 0.4726 and θ = 28.20 o The angle between the two rays is the difference so = 0.34 o . [NB with a prism red is deviated least (“red tries to go straight ahead” pronounce ‘ahead’ as “a red”), useful to learn. The diffraction grating spreads out the colours with blue deviated most.) n = C/V n 1 =1 n 2 =0.4674

10. Which way will the light refract? The rules Draw the normal line going through the prism (90 degrees from the surface) If light is going into something more dense it will bend towards the normal Draw a normal line for the light coming out of the prism Light will bend away from the normal if going into a less dense material

11. n = 1.5 n = 1.4 n =1.6 n = 1.5 n = 1.4 n =1.6 n = 1.5 n =1.4 n =1.6 n = 1.7 n = 1.7 n =1.7

12. Wave fact revision! Frequency f is determined by the source of the waves (the “transmitter”) Wave speed v is determined by the medium in or on which the wave propagates. So wavelength follows from v = fλ  At boundaries (reflection, refraction etc), f must stay the same.

13. On a rope

14. Superposition When two waves arrive at the same point and at the same time, the resultant displacement is given by the algebraic sum of the two individual displacements. (‘Algebraic sum’ means that you have to take account of positive and negative values.) + + = = Reinforcement Cancellation

15. Superposition of waves worksheet

16. Superposition in water Create your own wave superposition in your tub. Take a photograph and print off a picture Show on the picture where the waves have a supercrest, where they have a supertrough Where a crest meets a trough and the resultant displacement is zero

17. Paste pictures in here (at least 4 per page)