Chapter K - Problems

Transcription

1 Chapter K - Problem Blinn College - Phyic Terry Honan Problem K. A He-Ne (helium-neon) laer ha a wavelength of nm. If thi i hot at an incident angle of 55 into a gla block with index n =.52 then (a) what i the frequency of thi light, (b) what i the eed of the light in the gla, (c) what i the wavelength of the light in the gla and (d) what i the angle the beam make from the normal to the urface when inide the gla? Solution to K. (a) The wavelength i nm in a vacuum. The frequency i f = c l = 3.00µ µ0-9 = 4.74µ04 Hz (b) When light move from one medium to another it peed change v = c n = 3.00µ08.52 =.97µ0 8 m (c) Since the frequency of light doe not change and it peed doe it follow that the wavelength mut decreae by the ame factor a the peed. l = l 0 n = = 46 nm.52 (d) Take the index of air to be n =. Snell' law give the refracted angle. n in q = n 2 in q 2 ï in 55 =.52 in q 2 ï q 2 = 32.6 Problem K.2 When under water a wimmer ee the un make a 50 angle from the urface. What i the true angle of the un above the water' urface? Solution to K.2 The light move from the air to the water o n = and n 2 =.333. Since angle are alway meaured from the normal to the urface we have q 2 = = 40. n in q = n 2 in q 2 ï in q =.333 in 40 ï q 2 = 59.0 ï = 3.0

2 2 Chapter K - Problem Problem K.3 What i the critical angle for total internal reflection for a fiber optic tube with index.40? If the tube i ubmerged in water then what i the critical angle? Solution to K.3 tube in air ï in q crit = n 2 n =.40 ï q crit = 45.6 tube in water ï in q crit = n 2 n = ï q crit = 7.8 Problem K.4 Conider a clear thick flat horizontal urface with a mall imperfection.0 cm below the urface.when a penny i place directly over the imperfection it cannot be een at all but when a dime i placed over the imperfection it can be een by viewing from an angle. What doe thi imply about the index of refraction of the platic? A dime ha a radiu of cm and a penny ha a cm. Solution to K.4 The incident angle from the imperfection to the edge of the coin atifie: tan q = rêh, where h i the depth. q,dime = tan = and q,penny = tan = It follow that the critical angle i between the incident angle for the dime and penny. Thi give inequalitie for the index. q,dime q crit q,penny ï in q,dime in q crit = n in q,penny ï n ï.53 n.6 in q,penny in q,dime Problem K.5 The indice of refraction of red and violet light in a gla prim are.5 and.53, repectively. Suppoe the prim ha an apex angle of 55 and the light i incident at 40. What i the total angle of deflection for red, for violet and what i the total angle of diperion? Solution to K.5

3 Chapter K - Problem 3 For a given q and n we can find the other angle. To get q 2 ue Snell' law at the firt interface. in q = n in q 2 We can relate q 2 to q 2. Conider the triangle formed by the apex and the internal portion of the ray. The three internal angle of thi add to 80. At the econd interface we can then find q uing Snell' law. F + H90 - q 2 L + I90 - q 2 M = 80 ï q2 + q 2 = F. in q = n in q2 The total angle of diperion i d in the diagram. It i the total amount the ray bend due to the prim. At the firt interface it i deflected by q - q 2 at the econd it i deflected by q - q2. The total amount of deflection i the um of the two. d = q - q 2 + q - q2 ï d = q + q - F Applying thee expreion uing the number given give the following. F = 55 and q = 40 For red n =.5 For violet n =.53 in q = n in q 2 q 2 = q 2 = q 2 + q 2 = F q2 = q2 = in q = n in q2 q = q = d = q + q - F d = d = The total angle of diperion i the difference of thee two deflection angle d violet - d red =.59. Note that thi could alo be found by taking the difference between the value of q. Problem K.6 The minimum angle of deviation when a light beam i hot through a prim with a 45 apex angle i 3. What i the index of refraction for the prim? Solution to K.6 Refer to the previou problem. The angle of deflection d ha it minimum value d min when the incoming and outgoing ray are ymmetrical. Snell' law give the index. q = q and d = q + q - F ï q = d min + F = q 2 = q 2 and q 2 + q 2 = F ï q2 = F 2 = 45 2 = 22.5 = 38

4 4 Chapter K - Problem in q = n in q 2 ï n = in q in 38 = in q 2 in 22.5 =.6 Problem K.7 Conider a convex mirror with a 40 cm radiu. (a) If an object i 30 cm from the mirror then where i the image and what i it magnification? (b) Repeat (a) if the object i 60 cm from the mirror. Solution to K.7 For all cae of pherical mirror or lene we have For a convex mirror R < 0, o f = Rê2 = -40ê2 = -20 cm. (a) = 30 cm f = + and m = h = f O- = -2 cm m = - = = 0.4 (b) = 60 cm = f O- = -5 cm m = - = = 0.25 Problem K.8 A concave mirror ha a 60 cm radiu. (a) If a 5 cm high object i placed 90 cm from the mirror then where i the image and what i it height? (b) Repeat (a) if the ame object i placed 20 cm from the mirror. (c) Trace the ray for both cae. Solution to K.8 For a concave mirror f = R 2 = 60 2 (a) = 90 cm = 30 cm. f = + and m = h = f O- = 45 cm

5 Chapter K - Problem 5 h ï h 5 = = -0.5 ï h = -2.5 cm (b) = 20 cm = f O- = -60 cm h ï h 5 = = 3 ï h = 5 cm h = = 3 (c) To ray trace mark the center, a ditance R from the mirror, and mark the point at the poition of the object below the axi by the object' height. Problem K.9 When an object i 30 cm away from a pherical mirror an image i created that i inverted and four time larger than the object. What i the radiu of the mirror? I it concave or convex? Trace the ray for thi arrangement. Solution to K.9

6 6 Chapter K - Problem f = + and m = h The image ditance i = 30. The image i inverted o h i negative. -4 = h ï = 4 = 20 Since f and R are poitive it i a concave mirror. R 2 = f + O O- = 24 ï R = 48 cm Problem K.0 A concave make-up mirror ha a radiu R. To create an upright image that i four time larger than the object, where mut the object be placed? Solution to K.0 f = + and m = h The focal length i half the radiu, f = Rê2. The image i upright o it magnification i poitive. +4 = m = h ï = -4 2 R = f = + = + -4 ï = R 2 J - 4 N = 3 8 R Problem K. Arctic explorer ee a frozen animal that i 50 cm below the urface of a glacier. How far below the urface doe the animal appear when viewed from directly above? The index of refraction of ice i3. Solution to K. = - n 2 n ï 50 = -.3 ï = cm

7 Chapter K - Problem 7 The negative ign implie it i a virtual image. In thi cae it mean it i below the urface by 38.2 cm. Problem K.2 A gla (n =.50) phere with a 5 cm radiu ha a mall imperfection 5 cm above the center. When viewed from directly above where i the image? I it above or below the urface Solution to K.2 n + n 2 = n 2 -n R The ray i paing from inide the gla n =.50 to air n 2 =. The radiu i taken to be poitive when the center i on the ide where the light end up. In thi cae the center i where the light originate, o R = -5 cm. The object i = 5-5 = 0 cm from the urface = ï = cm It appear 8.57 cm below the urface. Problem K.3 A goldfih i inide a pherical bowl with a 2 cm radiu. Aume the gla of the bowl i thin o that the gla layer can be neglected. If a peron view the bowl with hi face at the ame level a the center of the bowl and the goldfih i on the ame radial line a the face. Where doe the goldfih ee the face if the face i 20 cm from the edge of the bowl? Where if the face i 5 cm from the bowl. Solution to K.3 n + n 2 = n 2 -n R The ray i paing from inide the air n = to water n 2 =.33. The radiu i taken to be poitive becaue the center i inide the bowl R = +2 cm. The negative q mean the face appear outide of the bowl = = 20 cm ï = -59. cm = 5 cm ï = -7.7 cm Problem K.4 What i the focal length of a biconvex len with urface of 0 cm and 5 cm radii, if the len i made of a platic material with n =.60? Would thi change i the len i revered? Solution to K.4

8 8 Chapter K - Problem Ue the lenmaker formula f = Hn - L -. The radiu i taken to be poitive if it center i on the ide where the light R R 2 emerge from the len. Since the len i biconvex the R > 0 and R 2 < 0. f = H.60 - L K 0 - O ï f = 0 cm -5 Revering the len make R = 5 cm and R 2 = -0 cm which give the ame value for f. Thi i a general feature of lene. Problem K.5 When an object i placed 0 cm from a thin len a virtual image i een 5 cm from the len. What i the focal length of the len? What type of lene i it, converging or diverging? Solution to K.5 Here we have = 0 cm and ince it i a virtual image = -5 cm. f = + ï f + O O- = 30 cm Problem K.6 A lide projector ue a converging len to project a 35 mm wide lide to fill a 70 cm wide creen. If the ditance from the lide to the creen i to be 6.3 m then where mut the lide be placed and what focal length mut the len have? Solution to K.6 h = 3.5 cm and h = -70 cm. Note that the image height i negative becaue it i inverted. Another way of eeing thi i the image mut be real becaue it i projected and a real image mut have a poitive q; thi require a negative h. The lide (object) to creen (image) ditance i +, - = h h = = -20 ï = 20 + = 630 cm ï + 20 = 630 cm ï = = 30 cm. = = 600 ï f + O O- = 28.6 cm. Problem K.7 An object i 20 cm from a diverging len with f = -32 cm. Where i the image and what i it magnification? Trace the ray for thi arrangement. Solution to K.7

9 Chapter K - Problem 9 f = + and m = h f = -32 cm and = 20 cm ï = f O- = -2.3 cm m = - = =