Solutions PHYS1252 Exam #1, Spring 2017, V hbs. Problem I: Multiple Choice Questions, 20P total = 5P(Q.1) + 5P(Q.2) + 5P(Q.3) + 5P(Q.

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1 Problem I: Multiple Choice Questions, 20P total = 5P(Q1) + 5P(Q2) + 5P(Q3) + 5P(Q4) Exam Version: 1A 1B 1C 1D 1E 1F Q1 [5P] B B C D C B Q2 [5P] D D C E A B Q3 [5P] A C C B B D Q4 [5P] D D E E A E Detailed solutions for very similar multiple choice questions are posted on the PHYS1212/PHYS1252 course website for: --Conceptual Practice Problems for Exam#1 --PHYS1112 Exam #1, Conceptual Problems, Spring 2009, 2010, 2011, PHYS1212 and PHYS1252 Exam#1, Problem I Questions, Spring In-class quizzes Q101-Q104 Also see: --LONCAPA Solutions for HW Sets #1 and #2

2 Problem II: Ray Diagram, 10P total Detailed Solution Steps for Exam Version #1A (see drawings below for all versions): 1P: Draw optical axis, mirror principal plane, H, and correctly placed image arrow to the left of the mirror, with all 3 elements properly labeled Image arrow can be chosen to point either upward or downward 1P: Correctly identify and label outgoing side of mirror, to the right of mirror Reasoning: i) image is virtual à by sign convention, image is located not on outgoing side ii) image is to the left of mirror à to the left of mirror is not the outgoing side iii) à outgoing side is to the right of the mirror 1P: Correctly identify and label incoming side, to the right of the mirror: Reasoning: i) The device is a mirror with reflected rays sent back to incident side, ie, leaving mirror on same to side they come in ii) à Incoming side is same as outgoing side: to the right of mirror 1P: Correctly place focal points F and F : F and F to the left of the mirror and to right of image on the optical axis Label F and F Reasoning: i) mirror is divergent, ie, has negative focal length f<0 ii) à by sign convention, F is not on in and F is not on out side, ie, F and F are both to the left of mirror iii) F and F are at both at a distance f from mirror à F and F coincide, at same distance from and on same side of the mirror iv) absolute image distance from mirror, d, is greater than absolute focal length, f v) à F and F are located to the right of the image, ie, between image and mirror 2P: Correctly draw one of the three principal ray pairs (see drawing Examples 1, 2): Either: F-F -ray Or: P-P -ray Or: C-C -ray 2P: Correctly draw a second principal ray pair (see drawing Examples 1, 2): Either: P-P -ray Or: C-C -ray Or: F-F -ray Note: For a mirror (unlike a mirror!) the C-ray and C -ray are two distinct lines, each intersecting the optical axis at the same angle, but one from above, the other from below So, the C -ray for a mirror would have to be constructed from the C-ray, by reflecting the C-ray at the optical axis That s unlike a lens, where the C-ray is simply continued straight through the mirror to get the C -ray 1P: Correctly draw object at intersection of two incoming principal rays, as arrow Object will be to the left of mirror Object arrow will be oriented opposite to image arrow, ie, inverted relative to image So, if image arrow was drawn upward, object arrow must point downward Else, if image arrow was drawn downward, object arrow must point upward Either choice of image arrow is acceptable 1P: State: object is virtual Reasoning: i) object is found to the right of the mirror à object is not on the incoming side ii) à by sign convention: object is virtual and d<0

3 Ray diagrams for all exam versions, #1A, 1B, 1C, 1D, 1E, 1F, are shown below In all versions, the object is virtual PHYS1252, Sp 2017, Exam #1, Problem II: Ray Diagram, Version #1A INPUTS: Image is virtual and to the left of mirror à Not Out = Left à Out = Right Device is a mirror (not a lens!) à In = Out, Not In = Not Out Mirror is divergent, f<0 à F on Not In, F on Not Out Not Out Not In Out In F-ray Q F -ray Image F F C-ray Optical Axis A Object P-ray Q RESULTS: Object is located on Not In à Object is virtual, d<0 Object and image arrows have opposite orientation àimage is inverted relative to object H P -ray C -ray PHYS1252, Sp 2017, Exam #1, Problem II: Ray Diagram, Version #1B F-ray Out In Not Out Not In INPUTS: Image is virtual and to the right of mirror à Not Out = Right à Out = Left Device is a mirror (not a lens!) à In = Out, Not In = Not Out Mirror is divergent, f<0 à F on Not In, F on Not Out F -ray Q C-ray F F Image Optical Axis A P-ray Object Q C -ray P -ray H RESULTS: Object is located on Not In à Object is virtual, d<0 Object and image arrows have opposite orientation àimage is inverted relative to object

4 PHYS1252, Sp 2017, Exam #1, Problem II: Ray Diagram, Version #1C H F-ray INPUTS: Image is real and to the right of mirror à Out = Right à Not Out = Left Device is a mirror (not a lens!) à In = Out, Not In = Not Out Mirror is divergent, f<0 à F on Not In, F on Not Out F -ray P -ray Q Image Q P-ray F F C -ray Optical Axis A Object RESULTS: Object is located on Not In à Object is virtual, d<0 Object and image arrows have same orientation àimage is erect relative to object Not Out Not In Out In C-ray PHYS1252, Sp 2017, Exam #1, Problem II: Ray Diagram, Version #1D F-ray H Q F -ray P -ray INPUTS: Image is real and to the left of mirror à Out = Left à Not Out = Right Device is a mirror (not a lens!) à In = Out, Not In = Not Out Mirror is divergent, f<0 à F on Not In, F on Not Out Image P-ray Q C -ray F F Object Optical Axis A C-ray Out Not Out In Not In RESULTS: Object is located on Not In à Object is virtual, d<0 Object and image arrows have same orientation àimage is erect relative to object

5 PHYS1252, Sp 2017, Exam #1, Problem II: Ray Diagram, Version #1E H F-ray F -ray INPUTS: Image is real and to the right of mirror à Out = Right à Not Out = Left Device is a mirror (not a lens!) à In = Out, Not In = Not Out Mirror is divergent, f<0 à F on Not In, F on Not Out P -ray Q Image Q P-ray F F C -ray Optical Axis A Object RESULTS: Object is located on Not In à Object is virtual, d<0 Object and image arrows have same orientation àimage is erect relative to object Not Out Not In Out In C-ray PHYS1252, Sp 2017, Exam #1, Problem II: Ray Diagram, Version #1F INPUTS: Image is virtual and to the left of mirror à Not Out = Left à Out = Right Device is a mirror (not a lens!) à In = Out, Not In = Not Out Mirror is divergent, f<0 à F on Not In, F on Not Out Not Out Not In Out In F-ray Q F -ray Image F F C-ray Optical Axis A Object P-ray Q RESULTS: Object is located on Not In à Object is virtual, d<0 Object and image arrows have opposite orientation àimage is inverted relative to object H P -ray C -ray

6 Problem III: Refraction at a Prism Immersed in Air, 35 P total = 15P(a) + 10P(b) + 10P(c) Detailed Solution for Exam Version #1A: Exercise 1 Surface L Surface R ϴ 1 O Glass n G Refraction at a prism immersed in air: Q α α Surface B Prism surface B and the incident ray (outside the prism) are horizontal The angle between prism surfaces L and B is α = 15 o The index of refraction of the glass is n G = 210 The refracted ray from surface L inside the prism strikes surface B a) Since incident ray (IR) is parallel to surface B (SB), the angle between IR and surface L (SL) is a=15 o That angle a and angle of incidence, Q1, add up to 90 o, since Q1 is measured from IR to the normal to surface L (NSL) Hence, the angle of incidence, Q 1, is [7P] Q1 = 90 o - a = 90 o - 15 o = 75 o By Snell s law of refraction: nair sin(q1) = ng sin(q2) and we can use nair» 10 à sin(q2) = (nair / ng ) sin(q1) = (10/210) sin(75 o ) = Hence, the angle of refraction, Q 2, is [8P] à Q2 = arcsin( ) = o b) Sum of interior angles in triangle OPQ is 180 o Hence, see drawing above: [8P] 180 o = (90 o + Q2 ) + f + a à f = 180 o - (90 o + Q2 + a) = 90 o - (27385 o + 15 o ) = o ϴ 2 φ c) That angle f and angle of incidence, Q3, add up to 90 o, since Q3 is measured from ray inside the prism (IP) to the normal of surface B (NSB) Hence: [5P] Q3 = 90 o - f = 90 o o = o The critical angle of incidence at surface B is: sin(qcrit) = (nair / ng ) = (10/210) = [5P] à Qcrit = arsin(nair / ng ) = arcsin(047619) = o à Q3 > Qcrit à Actual angle of incidence at SB exceeds critical angle of incidence à The ray incident at SB will undergo total internal reflection and no refracted ray emerges through SB from the prism into the air Alternatively, we could also use Snell s law to calculate the sine of the angle of refraction at SB (call it Q4, not shown in drawing), assuming there was a refracted ray: [10P-Alt] sin(q4) = (ng / nair ) sin(q3) = (21/10) sin(42385 o ) = 1416 > 1 Since sin(q4) exceeds 1, the angle of refraction, Q4, does not exist and no refracted ray exits at SB ϴ 3 P

7 Detailed Solution for Exam Version #1D: Exercise 2 ϴ 1 F Ɣ H Surface R α ϴ 2 ψ G Surface L α Glass n G ϴ 5 Refraction at a prism immersed in air: Surface B Prism surface B and the incident ray (outside the prism) are horizontal The angle between prism surfaces L and B is α = 35 o The apex angle between prism surfaces L and R is γ = 97 o The index of refraction of the glass is n G = 160 The refracted ray from surface L inside the prism strikes surface B a) Since incident ray (IR) is parallel to surface B (SB), the angle between IR and surface L (SL) is a=19 o That angle a and angle of incidence, Q1, add up to 90 o, since Q1 is measured from IR to the normal to surface L (NSL) Hence: [7P] Q1 = 90 o - a = 90 o - 35 o = 55 o By Snell s law of refraction: nair sin(q1) = ng sin(q2) and we can use nair» 10 à sin(q2) = (nair / ng ) sin(q1) = (10/160) sin(55 o ) = [8P] à Q2 = arcsin( ) = o b) Sum of interior angles in triangle FGH is 180 o Hence, see drawing above: [8P] 180 o = (90 o - Q2 ) + y + g à y = 180 o - (90 o - Q2 + g) = 90 o + Q2 - g y = 90 o o - 97 o = o c) That angle y and angle of incidence, Q5, add up to 90 o, since Q5 is measured from ray inside the prism (IP) to the normal of surface R (NSR) Hence: Q5 = 90 o - y [5P] Q5 = 90 o o = o The critical angle of incidence at surface R (SR) is: sin(qcrit) = (nair / ng ) = (10/16) = [5P] à Qcrit = arcsin( ) = o à Q5 > Qcrit à Actual angle of incidence at SR exceeds critical angle of incidence à The ray incident at SR will undergo total internal reflection and no refracted ray emerges through SR from the prism into the air Alternatively, we could also use Snell s law to calculate the sine of the angle of refraction at SR (call it Q6, not shown in drawing), assuming there was a refracted ray: [10P-Alt] sin(q6) = (ng / nair ) sin(q5) = (16/10) sin(66205 o ) = 1464 > 1 Since sin(q6) exceeds 1, the angle of refraction, Q6, does not exist and no refracted ray exits at SR

8 Numerical Solutions for Exam Versions #1A #1F: PHYS1252, Sp'17, Exam#1, Problems III Manual Inputs: Final Results: Part Exam#: 1A 1B 1C 1D 1E 1F Problem III Inputs: (a), (b) α [deg] (b), (c) ɣ [deg] N/A N/A N/A (a), (c) n_g Problem III Results: (a) θ_1 [deg] (a) sin(θ_2) (a) θ_2 [deg] (b) ϕ [deg] N/A N/A N/A (b) ψ [deg] N/A N/A N/A (c) θ_crit [deg] (c) θ_3 [deg] N/A N/A N/A (c) sin(θ_4) N/A N/A N/A (c) θ_4 [deg] N/A N/A N/A (c) Refracted ray thru B? No No No N/A N/A N/A (c) θ_5 [deg] N/A N/A N/A (c) sin(θ_6) N/A N/A N/A (c) θ_6 [deg] N/A N/A N/A (c) Refracted ray thru R? N/A N/A N/A No No No

9 Problem IV: Microscope or Telescope 35 P total = 15P(a) + 10P(b) + 10P(c) Detailed Solution for Exam Version #1A: Compound Microscope Compound Microscope: Initial Drawing 1: in 1: not out 1: not in 1: out L 2: in 2: not out 2: not in 2: out d 1 d 1 h 1 F 2 F 1 h 1 =h 2 d 2 Eye h 2 H 1 H 2 d 2 a) Find image of Lens 1: Img1 [1P] Given d1 =115cm and Obj1 is to the left = incoming side of Lens 1 à d1=+115cm > 0 Also given: f1=+110cm [8P] à d1 = (1/ f1-1/d1) -1 = (1/ 110-1/115) -1 cm = cm = distance from Img1 to Lens 1 d1 > 0 à Img1 is on outgoing side of Lens 1 à Img1 is to the right of Lens 1 d1 > 0 à Img1 is real m1 = - d1 / d1 = < 0 à Img1 is inverted rel to Obj1 (Note: only the correct sign of m1 is required here to get full credit)

10 b) Find Object of Lens 2: Obj2 Find Lens-to-Lens Distance [4P] Given d2 =29cm and Img2 is to the left = not outgoing side of Lens 2 àimg2 is virtual and d2 = - 29cm < 0 [4P] 2 Also given: f2=+190cm à d2 = (1/ f2-1/d2 ) -1 = (1/ 190-1/(-29)) -1 cm = cm = distance from Obj2=Img1 to Lens L = d1 + d2 = ( ) cm = 27083cm = distance Lens 2 from Lens 1 Notice also (not required, see drawing): d2 > 0 à Obj2 is on incoming side of Lens 1 à Obj2 is to the left of Lens 2 d2 > 0 à Obj2 is real m2 = - d2 / d2 = > 0 à Img2 is erect rel to Obj2 and Img2 is inverted rel to Obj1 c) Find Size of final image, Img2, and its orientation rel to original object, Obj1 m1 = - d1 / d1 = - (2530 / 115) = Given: h1 = 0095 mm à h1 = m1 h1 = (-2200) (0095 mm) = -209 mm m2 = - d2 / d2 = - ((-2900) / 17832) = h2 = h1 = -209 mm à h2 = m2 h2 = (+1626) (-209 mm) = -340 mm; final image diameter is 340mm m1 < 0 à Img1 inverted rel to Obj1 m2 > 0 à Img2 erect rel to Obj2=Img1 àimg2 inverted rel to Obj1 Alternative solution: [6Palt] mtot = m1 m2 = (d1 / d1) (d2 / d2 ) = (2530/ 115) ((-2900)/ 17832) = Given: h1 = 0095 mm [2Palt] à h2 = mtot h1 = (-3578) (0095 mm) = -340 mm; final image diameter is 340mm [2Palt] mtot < 0 à Img2 inverted rel to Obj1

11 Detailed Solution for Exam Version #1D: Galilean Telescope Galilean Telescope: Initial Drawing Convergent Lens 1: f 1 >0 Divergent Lens 2: f 2 <0 Obj1 In, Not Out Out, Not In In, Not Out Out, Not In H 1 L H 2 h 1 Img2 Img1 = Obj2 h 2 ' F 2 Eye F 2 F 1 h 1 =h 2 d 2 d 2 d 1 d 1 a) Find image of Lens 1: Img1 [1P] Given d1 = km = cm is much, much larger than given f1=+1120cmand Obj1 is to the left = incoming side of Lens 1 à 1/d1 is very, very small compared to 1/ f1 à 1/d1 can be neglected in the image formation eq for d1 : [8P] à d1 = (1/ f1-1/d1) -1 (1/ f1) -1 = f1 = cm = distance from Img1 to Lens 1 d1 > 0 à Img1 is on outgoing side of Lens 1 à Img1 is to the right of Lens 1 d1 > 0 à Img1 is real Given Obj1 is to the left = incoming side of Lens 1 à d1 = cm > 0 à m1 = - d1 / d1 = < 0 à Img1 is inverted rel to Obj1 (Note: only the correct sign of m1 is required here to get full credit)

12 b) Find Image of Lens 2: Img2 Given lens-to-lens distance L=1087cm and L = d1 + d2 à d2 = L - d1 = ( ) cm = - 33cm Also given: f2 = -30cm [4P] à d2 = (1/ f2-1/d2) -1 = (1/(-30) - 1/(-33)) -1 cm = cm; distance from Img2 to Lens 2 is 330cm d2 < 0 à Img2 is on not on outgoing side of Lens 2 à Img2 is to the left of Lens 2 d2 < 0 à Img2 is virtual Notice also (not required, see drawing): d2 < 0 à Obj2 is not on incoming side of Lens 1 à Obj2 is to the right of Lens 2 d2 < 0 à Obj2 is virtual m2 = - d2 / d2 = < 0 à Img2 is inverted rel to Obj2 and Img2 is erect rel to Obj1 c) Find Size of final image, Img2, and its orientation rel to original object, Obj1 à m1 = - d1 / d1 = - (+112 / ) = < 0 Given: h1 = km = mm à h1 = m1 h1 = ( ) ( mm) = mm m2 = - d2 / d2 = - ((-3300)/ (-33)) = h2 = h1 = mm à h2 = m2 h2 = (-1000) ( mm) = +213 mm; final image diameter is 213 mm m1 < 0 à Img1 inverted rel to Obj1 m2 < 0 à Img2 inverted rel to Obj2=Img1 àimg2 erect rel to Obj1 Alternative solution: [6Palt] mtot = m1 m2 = (d1 / d1) (d2 / d2 ) = (+112 / ) ((-3300)/ (-33)) = > 0 Given: h1 = km = mm [2Palt] à h2 = mtot h1 = ( ) ( mm) = +212 mm; final image diameter is 212 mm [2Palt] mtot > 0 à Img2 erect rel to Obj1

13 Problem V: Angular Magnification by Microscope or Telescope 10P (Bonus) Detailed Solution for Exam Version #1A: Angular Magnification by Compound Microscope Using inputs and results from Problem IV, Part (c) Compound Microscope: Angular Magnification M Θ = θ e / θ ref Img2 θ e h e = h 2 d 2 =d e h ref = h 1 θ ref Obj1 d ref = d near [4P] [4P] From Problem IV, input: de = d2 = 290 cm From Problem IV, Part (c): he = h2 = 340 mm = 340 cm à θe tan( θe ) = he / de = 340 /290 = radians = 672 o Or: θe = arctan( he / de ) = arctan(340 /290) = radians = 669 o Given: dref = dnear = 250 cm From Problem IV, input: href = h1 = 0095 mm = cm à θref tan( θref ) = href / dref = /250 = radians = o Or: θref = arctan( href / dref ) = arctan(340 /290) = radians = o It is sufficient if θe and θref are both stated only in radians or only in degrees à M ' = θe / θref = (01172) / ( ) = 308 Or: M ' = θe / θref = (01167) / ( ) = 307 Will also accept M ' = 308 or M ' = 307 as correct solutions, with M ' < 0 indicating that the final image seen by the eye (Img2) is inverted relative to the original object (Obj1)

14 Detailed Solution for Exam Version #1D: Angular Magnification by Galilean Telescope Using inputs and results from Problem IV, Parts (b) and (c) Galilean Telescope: Angular Magnification M Θ = θ e / θ ref d 2 =d e h e = h 2 θ e Img2 h ref = h 1 θ ref Obj1 d ref = d 1 [4P] [4P] From Problem IV, Part (b): de = d2 = 330 cm From Problem IV, Part (c): he = h2 = 212 mm = 0212 cm à θe tan( θe ) = he / de = 0212 /330 = radians = o Or: θe = arctan( he / de ) = arctan(0212 /330) = radians = o From Problem IV, input: dref = d1 = km From Problem IV, input: href = h1 = km à θref tan( θref ) = href / dref = ( km) /( ) = radians = o Or: θref = arctan( href / dref ) = arctan(( km) /( )) = radians = o It is sufficient if θe and θref are both stated only in radians or only in degrees à M ' = θe / θref = ( ) / ( ) = 339

15 Numerical Solutions for Exam Versions #1A #1F: PHYS1252, Sp'17, Exam#1, Problems IV and V Manual Inputs: Final Results: Part Exam#: 1A 1B 1C 1D 1E 1F Problem IV Inputs: (a) f_1 [cm] (b) f_2 [cm] (a),(c) d_1 [cm] E E E+13 (b) L [cm] N/A N/A N/A (b),(c) d_2' [cm] N/A N/A N/A (b) Img2 rel to Lens2 left of left of left of N/A N/A N/A (c) h_1 [cm] E E E+10 Problem V Inputs: d_near [cm] N/A N/A N/A Problem IV Results: (a) d_1' [cm] (a) Img1 rel to Lens1 right of right of right of right of right of right of (a) Img1 real or virtual real real real real real real (a) sign(m_1) (a) Img1 orientn rel to Obj1 inverted inverted inverted inverted inverted inverted (b) d_2 [cm] (b) d_2' [cm] (b) Img2 rel to Lens2 N/A N/A N/A left of left of left of (b) Img2 real or virtual virtual virtual virtual virtual virtual virtual (b) L [cm] N/A N/A N/A (c) m_ E E E-12 (c) m_ (c) m_tot E E E-11 (c) h_1'=h_2 [cm] (c) h_2' [cm] (c) Img2 orientn rel to Obj1 inverted inverted inverted erect erect erect Problem V Results: h_e [cm] E E E-01 d_e [cm] E E E+01 θ_e [radians] E E E-03 θ_e [degrees] E E E-01 h_ref [cm] E E E+10 d_ref [cm] E E E+13 θ_ref [radians] E E E E E E-04 θ_ref [degrees] E E E E E E-02 M_θ