LECTURE ON NUMERICAL PROBLEMS IN OPTICS

1. LECTURE ON NUMERICAL PROBLEMS IN OPTICS BY KAVITA MONGA(LECTURER-PHYSICS) GOVT POLYTECHNIC COLLEGE KHUNIMAJRA, MOHALI DATE- 18/4/2013

2. Light :- i) Light is a form of energy which helps us to see objects. ii) When light falls on objects, it reflects the light and when the reflected light reaches our eyes then we see the objects. iii) Light travels in straight line. iv) The common phenomena of light are formation of shadows, formation of images by mirrors and lenses, bending of light by a medium, twinkling of stars, formation of rainbow etc.

4. The overall study of how light behaves is called optics. The branch of optics that focuses on the creation of images is called geometric optics, because it is based on relationships between angles and lines that describe light rays.

5. Some Definitions • Absorption • When light passes through an object the intensity is reduced depending upon the color absorbed. Thus the selective absorption of white light produces colored light. • Refraction • Direction change of a ray of light passing from one transparent medium to another with different optical density. A ray from less to more dense medium is bent perpendicular to the surface, with greater deviation for shorter wavelengths • Diffraction • Light rays bend around edges - new wavefronts are generated at sharp edges - the smaller the aperture the lower the definition • Dispersion • Separation of light into its constituent wavelengths when entering a transparent medium - the change of refractive index with wavelength, such as the spectrum produced by a prism or a rainbow

6. Absorption Control Absorption No blue/green light red filter

7. A lens is an optical device that is used to bend light in a specific way. A converging lens bends light so that the light rays come together to a point. A diverging lens bends light so it spreads light apart instead of coming together.

8. Mirrors reflect light and allow us to see ourselves. A prism is another optical device that can cause light to change directions. A prism is a solid piece of glass with flat polished surfaces.

9. Optical Systems • An optical system is a collection of mirrors, lenses, prisms, or other optical elements that performs a useful function with light. • Characteristics of optical systems are: • The location, type, and magnification of the image. • The amount of light that is collected. • The accuracy of the image in terms of sharpness, color, and distortion. • The ability to change the image, like a telephoto lens on a camera. • The ability to record the image on film or electronically.

10. Images appear in mirrors because of how light is reflected by mirrors. The incident ray follows the light falling onto the mirror. The reflected ray follows the light bouncing off the mirror. Reflection

11. In specular reflection each incident ray bounces off in a single direction. A surface that is not shiny creates diffuse reflection. In diffuse reflection, a single ray of light scatters into many directions. Reflection

12. Law of Reflection The incident ray strikes the mirror. The reflected ray bounces off. The angle of incidence equals the angle of reflection.

13. A light ray is incident on a plane mirror with a 30 degree angle of incidence. Sketch the incident and reflected rays and determine the angle of reflection. Law of reflection 30o 30o

14. A dentist uses a mirror to look at the back of a second molar (A). Next, she wishes to look at the back of a lateral incisor (B), which is 90° away. By what angle should she rotate her mirror? Mirror A B A. 90° B. 45° C. 180°

15. Specular vs. Diffuse Reflection • Specular Reflection • The surface is flat at distance scales near or above the wavelength of light • It looks “shiny”, like a mirror.

16. Specular vs. Diffuse Reflection • Diffuse Reflection • The surface is rough at distance scales near or above the wavelength of light • Almost all surfaces reflect in this way!

17. Two plane mirrors form a right angle. How many images of the ball can you see in the mirrors? 1 2 3 4

18. Light rays may bend as they cross a boundary from one material to another, like from air to water. This bending of light rays is known as refraction. The light rays from the straw are refracted (or bent) when they cross from water back into air before reaching your eyes. Refraction

19. When a ray of light crosses from one material to another, the amount it bends depends on the difference in index of refraction between the two materials. Refraction

20. Air qA Water qw refraction Refraction Refraction is the bending of light as it passes from one medium into another. N Note: the angle of incidence qAin air and the angle of refraction qAin water are each measured with the normal N. The incident and refracted rays lie in the same plane and are reversible.

22. A ray of light traveling through air is incident on a smooth surface of water at an angle of 30° to the normal. Calculate the angle of refraction for the ray as it enters the water. Calculate the angle of refraction

23. solution • 1) You are asked for the angle of refraction. • 2) You are told the ray goes from air into water at 30 degrees. • 3) Snell’s law: ni sin(θi) = nr sin(θr) ni = 1.00 (air), nr = 1.33 (water) • 4) Apply Snell’s law to find θr. 1.00sin(30°) = 1.33 sin(θr) sin(θr) = 0.5 ÷ 1.33 = 0.376 • Use the inverse sine function to find the angle that has a sine of 0.376. • θr = sin-1(0.376) = 22°

24. Air Air Water Water Refraction Distorts Vision The eye, believing that light travels in straight lines, sees objects closer to the surface due to refraction. Such distortions are common.

25. Refraction He sees the fish here…. But it is really here!!

26. Depth perception

27. Apparent Depth

28. Virtual Image of Fish air water A fish swims below the surface of the water. An observer sees the fish at: A. a greater depth than it really is. B. its true depth. C. a smaller depth than it really is.

29. air water A fish swims directly below the surface of the water. An observer sees the fish at: A. a greater depth than it really is. B. its true depth. C. a smaller depth than it really is.

30. c Index of refraction v The Index of Refraction The index of refraction for a material is the ratio of the velocity of light in a vacuum (3 x 108 m/s) to the velocity through the material. Examples: Air n= 1; glass n = 1.5; Water n = 1.33

31. Index of Refraction • vmedium is the speed of light in a transparent medium. • c is the speed of light in a vacuum (c=3.00×108 m/s) • n is a dimensionless constant: n≥1 • n=1 in a vacuum

32. The ability of a material to bend rays of light is described by the index of refraction (n). Index of refraction

33. vair= c Air Glass n = 1.50 For glass: vG= 2 x 108 m/s Example 1.Light travels from air (n = 1) into glass, where its velocity reduces to only 2 x 108 m/s.What is the index of refraction for glass? If the medium were water: nW = 1.33. Then you should show that the velocity in water would be reduced from c to 2.26 x 108 m/s.

34. Pavement Air Glass Sand Analogy for Refraction 3 x 108 m/s 2 x 108 m/s vs < vp 3 x 108 m/s Light bends into glass then returns along original path much as a rolling axle would when encountering a strip of mud.

35. Snell's law of refraction • Snell’s law is the relationship between the angles of incidence and refraction and the index of refraction of both materials. Angle of refraction (degrees) Angle of incidence (degrees) nisinQi= nr sin Qr Index of refraction of incident material Index of refraction of refractive material

36. Snell’s Law of Refraction

37. Medium 1 q1 v1 v2 q2 Medium 2 Snell’s Law: Snell’s Law The ratio of the sine of the angle of incidence q1 to the sine of the angle of refraction q2 is equal to the ratio of the incident velocity v1to the refracted velocity v2 .

38. Air 300 H2O qW Example 2:A laser beam in a darkened room strikes the surface of water at an angle of 300. The velocity in water is 2.26 x 108 m/s. What is the angle of refraction? The incident angle is: qA qA = 900 – 300 = 600 qW = 35.30

39. Medium 1 q1 q2 Medium 2 Snell’s Law and Refractive Index Another form of Snell’s law can be derived from the definition of the index of refraction: Snell’s law for velocities and indices:

40. A Simplified Form of the Law Since the indices of refraction for many common substances are usually available, Snell’s law is often written in the following manner: The product of the index of refraction and the sine of the angle is the same in the refracted medium as for the incident medium.

41. Air Glass q qG 500 n=1.5 Air qe = 500 Apply to each interface: Same as entrance angle! Example 3. Light travels through a block of glass, then remerges into air. Find angle of emergence for given information. First find qG inside glass: qG = 30.70 qG From geometry, note angle qG same for next interface.

42. Air n=1 Glass n=1.5 lA lG Wavelength and Refraction The energy of light is determined by the frequency of the EM waves, which remains constant as light passes into and out of a medium. (Recall v = fl.) fA= fG lG < lA

43. The Many Forms of Snell’s Law: Refraction is affected by the index of refraction, the velocity, and the wavelength. In general: Snell’s Law: All the ratios are equal. It is helpful to recognize that only the index n differs in the ratio order.

44. Air Glass q qG qG n=1.5 q Air Example 4:A helium neon laser emits a beam of wavelength 632 nm in air (nA = 1). What is the wavelength inside a slab of glass (nG = 1.5)? nG = 1.5; lA = 632 nm Note that the light, if seen inside the glass, would be blue. Of course it still appears red because it returns to air before striking the eye.

45. Air 900 qc Critical angle Water light Total Internal Reflection When light passes at an angle from a medium of higher index to one of lower index, the emerging ray bends away from the normal. When the angle reaches a certain maximum, it will be reflected internally. The critical angle qc is the limiting angle of incidence in a denser medium that results in an angle of refraction equal to 900. i = r

47. Total Internal Reflection • Occurs when n2<n1 • θc= critical angle. • When θ1≥θc, no light is transmitted through the boundary; 100% reflection

48. Critical angle Air 900 qc Critical angle: qc = 48.80 Water Example 5.Find the critical angle of incidence from water to air. For critical angle, qA = 900 nA= 1.0; nW = 1.33 In general, for media where n1 > n2 we find that:

49. ref rac tion Refraction & Dispersion Short wavelengths are “bent” more than long wavelengths dispersion Light is “bent” and the resultant colors separate (dispersion). Red is least refracted, violet most refracted.

50. Red Orange Yellow Green Blue Indigo Violet Dispersion by a Prism Dispersion is the separation of white light into its various spectral components. The colors are refracted at different angles due to the different indexes of refraction.