Solutions PHYS1252 Exam #1, Spring 2016, 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 Solutions PHYS1252 Exam #1, Spring 2016, V hbs 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] C E A A C E Q2 [5P] B A D D B A Q3 [5P] E C C C E C Q4 [5P] D B C C D B 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 2015 Also see: - - LONCAPA Solutions for HW Sets #1 and #2

2 Solutions PHYS1252 Exam #1, Spring 2016, V hbs Problem II: Ray Diagram, 10P total Detailed Solution Steps for Exam Version #1A (see drawings below for all versions): 1P: Draw optical axis, lens principal plane, H, and correctly placed image as arrow to the left of the lens, 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 lens, to the right of lens Reasoning: i) image is virtual by sign convention, image is located not on outgoing side ii) image is to the left of lens to the left of lens is not the outgoing side iii) outgoing side is to the right of the lens 1P: Correctly identify and label incoming side, to the left of the lens: Reasoning: i) The device is a lens with refracted rays passing through, ie, leaving opposite to side they entered ii) Incoming side is opposite to outgoing side: to the left of lens 1P: Correctly place focal points F and F : F to the left of the lens, and right of image on the optical axis; F at same distance to the right of lens Label F and F Reasoning: i) lens is divergent, ie, has negative focal length f<0 ii) by sign convention, F is not on in and F is not on out side, ie, F is to the left and F to the right of lens iii) F and F are at both at a distance f from lens F and F are at same distance from, and on opposite sides of the lens iv) absolute image distance from lens, d, is greater than absolute focal length, f v) F is located to the right of the image, ie, between image and lens 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 lens!) the C- ray and C - ray would be 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 lens to get the C - ray] 1P: Correctly draw object at intersection of two incoming principal rays, as arrow, to the left of lens, arrow 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 lens object is not on the incoming side ii) by sign convention: object is virtual and d<0

3 Solutions PHYS1252 Exam #1, Spring 2016, V hbs Ray diagrams for all exam versions, #1A, 1B, 1C, 1D, 1E, 1F, are shown below In all versions, the object is virtual PHYS1252,)Sp 2016,)Exam)#1,)Problem)II:)Ray)Diagram,)Version)#1A) )Divergent)Lens:)F)not)on)in,)F )not)on)out) ) Not)Out )Out) ))))))))))In )Not)In) FDray) Q ) F Dray) Image) P Dray) CD,)C Dray) F ) )F) OpMcal)Axis)A) PDray) Q) Object) H) P Dray) PHYS1252,)Sp 2016,)Exam)#1,)Problem)II:)Ray)Diagram,)Version)#1B) )Divergent)Lens:)F)not)on)in,)F )not)on)out) ) Not)Out )Out) ))))))))))In )Not)In) FDray) F Dray) Q ) PDray) P Dray) Q) Image) Object) F ) F) OpMcal)Axis)A) CD,)C Dray) H)

4 Solutions PHYS1252 Exam #1, Spring 2016, V hbs PHYS1252,)Sp 2016,)Exam)#1,)Problem)II:)Ray)Diagram,)Version)#1C) Convergent)Lens:)F)on)in,)F )on)out) ) Not)Out )Out) ))))))))))In )Not)In) PDray) Q) P Dray) Object) Q ) F Dray) FDray) Image) F) F ) CD,)C Dray) OpMcal)Axis)A) PHYS1252,+Sp 2016,+Exam+#1,+Problem+II:+Ray+Diagram,+Version+#1D+ Convergent)Lens:)F)on)in,)F )on)out) + Not+Out +Out In +Not+In+ P'ray+ Q+ P 'ray+ Object+ Q + F 'ray+ F'ray+ Image+ F+ F + C',+C 'ray+ Op9cal+Axis+A+

5 Solutions PHYS1252 Exam #1, Spring 2016, V hbs PHYS1252,+Sp 2016,+Exam+#1,+Problem+II:+Ray+Diagram,+Version+#1E+ +Divergent)Lens:)F)not)on)in,)F )not)on)out) + Not+Out +Out In +Not+In+ F'ray+ Q + F 'ray+ Image+ P 'ray+ C',+C 'ray+ F + +F+ Op9cal+Axis+A+ P'ray+ Q+ Object+ H+ P 'ray+ PHYS1252,+Sp 2016,+Exam+#1,+Problem+II:+Ray+Diagram,+Version+#1F+ +Divergent)Lens:)F)not)on)in,)F )not)on)out) + Not+Out +Out In +Not+In+ F'ray+ F 'ray+ Q + P'ray+ P 'ray+ Q+ Image+ Object+ F + F+ Op9cal+Axis+A+ C',+C 'ray+ H+

6 Solutions PHYS1252 Exam #1, Spring 2016, V hbs Problem III: Refraction at a Glass Wall, 35 P total = 15P(a) + 10P(b) + 10P(c) Detailed Solution for Exam Version #1A: Solu2on:" x" beam"2" beam"1" θ" α" φ" θ " beam"1" y" φ " beam"2" (a) [15P] α = 59 o [5P] Left wall, beam 2 angle of incidence = θ= 90 o - α = 90 o - 59 o = 31 o [5P] Left wall, beam 2 angle of refraction = θ, tan(θ ) = y/x θ = arctan(y/x) = arctan( (125cm)/(36cm) )= o nwater = 1333 Snell s Law for beam 2 at left wall: nwater sin(θ) = nglass sin(θ ) [5P] nglass = nwater sin(θ) / sin(θ ) = (1333) sin(31 o ) / sin(19148 o ) = 2093 (b) [10P] [3P] Right wall, beam 2 angle of incidence = ϕ = θ = o [3P] [4P] Right wall, critical angle of incidence = ϕcrit : sin(ϕcrit) = nair / nglass = 100 / 2093 = ϕcrit = 2854 o ϕ < ϕcrit No TIR of beam 2 at the right wall Alternative Route: show that sin(ϕ ) < 1 (see below) No TIR of beam 2 at the right wall Right wall, beam 2 angle of refraction = ϕ Snell s Law for beam 2 at left wall: nglass sin(ϕ) = nair sin(ϕ ) sin(ϕ ) = (nglass / nair ) sin(ϕ) = (2093 / 100) sin(19148 o ) = sin(ϕ ) < 1 angle of refraction, ϕ, does exist; ϕ = arcsin(06865) = 4334 o

7 Solutions PHYS1252 Exam #1, Spring 2016, V hbs (c) [10P] nglass = 2093 c = m/s (vacuum speed of light) [4P] Speed of wave propagation inside the glass wall: vglass= c / nglass = ( m/s) / (2093) = m/s Beam 2 travel distance inside the glass wall (using Pythagoras, see drawing above): [4P] d= [ x 2 +y 2 ] 1/2 = [ (360) 2 +(125) 2 ] 1/2 cm = 3811cm = 03811m [2P] Travel time of beam 2 (crests or troughs) through the wall: t = d/ vglass = (03811 m) / ( m/s) = s = 266 ns

8 Solutions PHYS1252 Exam #1, Spring 2016, V hbs Problem IV: Microscope Image, 35 P total = 10P(a) + 15P(b) + 10P(c) Detailed Solution for Exam Version #1A: (a) [10P] Draw Obj1, Lens 1 and Lens 2 Find In, Out, Not In, Not Out Then draw F1, and d1: Compound$Microscope:$Step3$ Compound$Microscope:$Step5$ General$Direc>on$of$Light$Rays$ Passing$thru$Lenses$ d 1 $ F 1 $ [05P] d1 = +105cm and d1 > 0, since Obj1 (Neuron) is on In rel to Lens 1 (see drawing above) [05P] f1 = +100cm d1 = (1/ f1-1/ d1 ) - 1 = [ 1/(+100) - 1/(+105)] - 1 cm = +210cm [3P] Since d1 =210cm, the image by Lens 1, Img1, is 210cm from Lens 1 [3P] Since d1 >0, Img1, is on Out rel to Lens 1, ie, to the right of Lens 1 (see drawing above&below) [1P] Since d1 >0, the Img1 is real [2P] m1 = - (d1 / d1) = - [(+210cm)/(+105cm) = - 20 Since m1 < 0, Img1 is inverted rel to Obj1 Compound$Microscope:$Step6$ d 1 $ d 1 # F 1 $

9 Solutions PHYS1252 Exam #1, Spring 2016, V hbs (b) [15P] d2 =30cm [5P] d2 = - 30cm <0, since the image of Lens2, Img2, is to the left, ie, Not on Out of Lens2 (see left drawing below) [5P] [5P] d2 = (1/ f2-1/ d2 ) - 1 = [ 1/(+200) - 1/(- 300)] - 1 cm = +1875cm The object of Lens 2, Obj2, is 1875cm from Lens2 Note also: Since d2>0, Obj2, is on In rel to Lens 2, ie, to the left of Lens 1 (see right drawing below) L= d1 + d2 = (=210cm) + (+1875cm) =22875cm is the spacing between the two lenses (see right drawing below) Compound$Microscope:$Step7$ Compound$Microscope:$Step9$ L# d 1 $ d 1 # d 1 $ d 1 # F 1 $ F 1 $ F 2 $ d 2 $ d 2 $ d 2 $

10 Solutions PHYS1252 Exam #1, Spring 2016, V hbs (c) [10P] Route 1: h1 = 012mm = 0012cm m1 = - d1 / d1 = [see a), above] [3P] h1 = m1 h1 = (- 20) (0012 cm) = - 024cm < 0 [4P] [3P] h1 = h2 = - 024cm m2 = - d2 / d2 = - (- 30cm)/(+1875cm) = +160 > 0 h2 = m2 h2 = (+16) (- 024 cm) = - 384cm <0 As seen through Lens2, the absolute diameter of the neuron s final image, Img2, is 384cm Since h2 <0 and h1>0, final image, Img2, is inverted relative to Obj1 (=original object) Route 2: mtot = m1 m2 = (d1 / d1) (d2 / d2) = [(+210cm)/(+105cm)] [(- 300cm)/(+1875cm)] = h2 = mtot h1 = (- 320) (0012 cm) = - 384cm <0 [7P] As seen through Lens2, the absolute diameter of the neuron s final image, Img2, is 384cm [3P] Since mtot < 0, final image, Img2, is inverted relative to Obj1 (=original object) Compound$Microscope:$Step11$ L# d 1 $ d 1 # h 1 $ F 2 $ F 1 $ h 1 =h 2 $ d 2 $ h 2 $ d 2 $

11 Solutions PHYS1252 Exam #1, Spring 2016, V hbs Problem V: Microscope Angular Magnification, 10 P total (Bonus) Detailed Solution for Exam Version #1A: Compound$Microscope:$Step14$ M Θ $=$θ e$ /$θ ref $ θ e $ h 2 =h e # h O =h 1 $ d near $ d 2 =d e $ θ ref $ he = h2 = 384cm de = d2 = 300 cm [4P] θe tan(θe ) = he / de = = (384) / (300) = 0128 radian href = h1 = 012 mm = 0012 cm dref = dnear = 200 cm [4P] θref tan(θref ) = href / dref = (0012) / (200) = radian [2P] Mθ = θe / θref = (0128) / ( ) = 213