Presentation on Snells Law Snells Law Snells law

Snell’s Law Snell’s law states that a ray of light bends in such a

Snell’s Law Derivation 1 A n 1 x A n 2 2 • •

Snell’s Law Derivation A n 1 x A • n 2 2 (cont. )

$Refraction Problem #1 Goal: Find the angular displacement of the ray after having passed$

$Refraction Problem #2 Goal: Find the distance the light ray displaced due to the$

$Refraction Problem #3 Goal: Find the exit angle relative to the horizontal. = 19.$

Reflection Problem Goal: Find incident angle relative to horizontal so that reflected ray will

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Presentation on Snell’s Law

Snell’s Law Snell’s law states that a ray of light bends in such a way that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. Mathematically, i r ni nr ni sin i = nr sin r Here ni is the index of refraction in the original medium and nr is the index in the medium the light enters. i and r are the angles of incidence and refraction, respectively. Willebrord Snell

Snell’s Law Derivation 1 A n 1 x A n 2 2 • • d • B y • B Two parallel rays are shown. Points A and B are directly opposite one another. The top pair is at one point in time, and the bottom pair after time t. The dashed lines connecting the pairs are perpendicular to the rays. In time t, point A travels a distance x, while point B travels a distance y. sin 1 = x / d, so x = d sin 1 sin 2 = y / d, so y = d sin 2 Speed of A: v 1 = x / t Speed of B: v 2 = y / t Continued…

Snell’s Law Derivation A n 1 x A • n 2 2 (cont. ) 1 • d • B y • B v 1 x/ t x sin 1 = = = v 2 y/ t y sin 2 v 1 / c v 2 / c = sin 1 sin 2 1 / n 1 1 / n 2 = n 1 sin 1 = n 2 sin 1 sin 2 So, = n 2 n 1

$Refraction Problem #1 Goal: Find the angular displacement of the ray after having passed$

Refraction Problem #1 Goal: Find the angular displacement of the ray after having passed 1. Find the first angle of refraction through the prism. Hints: using Snell’s law. 19. 4712º 2. Find angle ø. (Hint: Use Geometry skills. ) 79. 4712º Air, n 1 = 1 30° Horiz. ray, parallel to base ø 3. Find the second angle of incidence. 10. 5288º 4. Find the second angle of refraction, , using Snell’s Law Glass, n 2 = 1. 5 15. 9º

$Refraction Problem #2 Goal: Find the distance the light ray displaced due to the$

Refraction Problem #2 Goal: Find the distance the light ray displaced due to the thick window and how much time it spends in the glass. Some hints are given. 20º 1 1. Find 1 (just for fun). 20º H 20 2. To show incoming & outgoing n 1 = 1. 3 rays are parallel, find . 20º 10 m d 0. 504 m glass 3. Find d. n 2 = 1. 5 4. Find the time the light spends in 5. 2 · 10 -8 s the glass. H 20 Extra practice: Find if bottom medium is replaced with air. 26. 4º

$Refraction Problem #3 Goal: Find the exit angle relative to the horizontal. = 19.$

Refraction Problem #3 Goal: Find the exit angle relative to the horizontal. = 19. 8° 36° air glass The triangle is isosceles. Incident ray is horizontal, parallel to the base. =?

Reflection Problem Goal: Find incident angle relative to horizontal so that reflected ray will be vertical. = 10º 50º center of semicircular mirror with horizontal base