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Magnification, Refraction and Snell’s Law

Magnification, Refraction and Snell’s Law. Warning – Lots of Math Required for this Lesson! SNC2D. Review Significant Digits. Non-zero digits are always significant! Zeroes placed before other digits are not significant; 0.046 has two significant digits

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Magnification, Refraction and Snell’s Law

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  1. Magnification, Refraction and Snell’s Law Warning – Lots of Math Required for this Lesson! SNC2D

  2. Review Significant Digits • Non-zero digits are always significant! • Zeroes placed before other digits are not significant; 0.046 has two significant digits • Zeroes placed between other digits are always significant; 4009 kg has four significant digits • Zeroes placed after other digits but before a decimal point are NOT significant; 7600 has two significant digits. • Zeroes placed after other digits but after a decimal point are significant; 7.90 has three significant digits

  3. Review of Scientific Notation! the decimal place is moved to the LEFT 4 places = 4.5 x 10-4 means 0.00045 • 4.5 x 10-4 means… • 5.6 x 105 means… the decimal place is moved to the RIGHT 5 places = 5.6 x 105means 560 000

  4. Converting to Scientific Notation • When writing in scientific notation we need to be aware of the significant digits. This will determine the first number in our term. • Example 1: 546 000 is written as 5.46 x 105 • Example 2: 0.0073 is written as 7.3 x 10-3 • Example 3: 5008 is written as 5.008 x 103

  5. Magnification • Magnification is the measure of how much larger or smaller an image is compared with the object itself. • It is expressed as a ratio of height of the image (hi) to the height of the object (ho) . • M = hi / ho • It can also be determined by taking the ratio of the distance from the image (di) to the mirror and the distance from the object (do)to the mirror • M = di / do

  6. Magnification Cont’d • Remember to use the same units of measurement in each ratio! • Your final answer will not include any units because they end up cancelling each other out. • If the Magnification factor is greater than 1, the image will be bigger than the object. • If the Magnification factor is less than 1, the image will be smaller than the object.

  7. Magnification Examples • Example 1: An object is placed 60 cm from a concave mirror. An image is produced 45 cm away. What is the magnification? • We are given distance… • M = di / do = 45 cm / 60 cm = 0.75 or 7.5 x 10-1 • Example 2: A concave mirror creates a virtual image of a candle flame that is 6 cm high. If the magnification of the mirror is 2.5, what is the height of the candle flame? • We are given height and magnification… • M = hi / ho rearrange to get the formula… ho = hi / M = 6 cm / 2.5 = 2.4 cm

  8. Refraction • Refraction - The bending of light rays as they pass through two different media (plural for medium) • Light travels very fast - 3.0 x 108 m/s in a vacuum • Light travels slower when it is moving through a medium (air, water, carbon dioxide, table salt, etc.) • Refraction occurs when light Enters a medium and also when it Leaves a medium.

  9. Index of Refraction • The amount by which a transparent medium decreases the speed of light is called refraction. • The larger the refractive index, the more the medium decreases the speed of light • Speed of light in a vacuum is 3.0 x 108 m/s. This is the fastest that light can travel, therefore it is given a Refractive Index of 1.00.

  10. Table 11.5 Index of Refraction Page 437 in your textbook!

  11. Index of Refraction Formula • Speed of light in a vacuum is denoted with a small letter c • Speed of light in a medium is denoted with a small letter v • The Refractive Index is denoted with a small letter n • The formula we use for index of refraction is n = c / v • In words: The index of refraction is equal to the speed of light in a vacuum divided by the speed of light in the medium

  12. Index of Refraction Problems • Example 1: The speed of light in leaded glass is 1.66 × 108 m/s. What is the index of refraction of this type of glass? n = c / v where c = speed of light in a vacuum = 3.0 x 108 / 1.66 x 108 = 3.0 x 108 m/s = 1.8v = 1.66 x 108 m/s (given) n = looking for Therefore, the index of refraction for leaded glass is 1.8.

  13. Index of Refraction Problems • Example 2: What is the speed of light through sapphire? n = c / v where c = speed of light in a vacuum 1.77 = 3.0 x 108 / v = 3.0 x 108 m/s 3.00 x 108 / 1.77 = v v = looking for 1.69 x 108 m/s = v n = 1.77 (table from pg. 435) Therefore, the speed of light through sapphire is 1.69 x 108 m/s

  14. Dispersion • As white light enters a water droplet, each wavelength of light gets refracted at slightly different angles. • The light gets refracted twice: • Once when it enters the water droplet • Once when it leaves

  15. Introduction to Snell’s Law If the light strikes the surface of the water at an angle, that part of the light beam that enters first will slow down first.

  16. Introduction to Snell’s Law 1.003 1.61

  17. Introduction to Snell’s Law • When light travels from air, with a low refractive index, into water, with a higher refractive index, it bends toward the normal • When light travels from a higher refractive index medium into a lower refractive index medium, it bends away from the normal • The angle of incidence, θi, and the angle of refraction, θR, are measured from the normal.

  18. Snell’s Law • Snell’s law is a formula that uses values for the index of refraction to calculate the new angle that a ray will take as a beam of light strikes the interface between two media: • n1sinθ1 = n2sinθ2 • The indices of refraction of two different media are indicated as n1 and n2 and the angles of incidence and the angle of refraction are indicated as θ1and θ2

  19. Snell’s Law Examples Solution: • Example 1: When light passes from air into water at an angle of 60° from the normal, what is the angle of refraction?

  20. Snell’s Law Examples • Example 2: In an experiment, a block of cubic zirconia is placed in water. A laser beam is passed from the water through the cubic zirconia. The angle of incidence is 50°, and the angle of refraction is 27°. What is the index of refraction of cubic zirconia? Solution:

  21. Seatwork Questions • Page 424 – 2 questions from EACH section • Page 425 – 2 questions from EACH section • Page 438 – All Practice Questions (except for the two we have already covered) • Page 441 – Practice Questions #1,2,3 • Page 442 – Practice Questions #1,2,3

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