# Physics 303L HW 11 - delgado (dd28287) hw11 hegelich...

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delgado (dd28287) – hw11 – hegelich – (55795)1Thisprint-outshouldhave28questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0pointsWhat must be the minimum length of a planemirror in order for you to see a full view ofyourself?1.One fourth of your heightWhat is the spatial length of this pulse as ittravels through glass with index of refractionn= 1.3?4.Twice of your height= 1.497×106mand the distance between the leading edge andthe rear edge of the pulse is1.497×106m,the required spatial length.003(part2of2)10.0pointsWhat is the spatial length of this pulse as ittravels through glass with index of refractionn= 1.3?
002(part1of2)10.0points
Laser scientists have succeeded in generatingpulses of light lasting as short as 1015s.Imagine that you create a pulse which lasts4.99×1015s.What is the spatial length of this pulsealong its direction of travel in a vacuum? Thespeed of light in a vacuum is 3×108m/s.00410.0pointsAn object, slanted at an angle of 45, isplaced in front of a vertical plane mirror, asshown.Mirror
Explanation:Which of the following shows the appar-ent position and orientation of the object’simage?
delgado (dd28287) – hw11 – hegelich – (55795)22.imagecorrect3.image4.image5.image6.image7.image8.imageExplanation:For a plane mirror, the line connecting apoint of the object and corresponding point ofthe image is perpendicular to the mirror, andthe corresponding distances from the pointsto the mirror are the same.Thus, the image should beimageobject00510.0pointsA coin is at the bottom of a beaker.Thebeaker is filled with 4.6 cm of water (n1=1.33) covered by 2.4 cm of liquid (n2= 1.4)floating on the water.How deep does the coin appear to be fromthe upper surface of the liquid (near the topof the beaker)?Correct answer: 5.17293 cm.Explanation:For small anglessinθtanθ=x.The appearance of the width of the coinxremains the same. Applying Snell’s Law,n1sinθ1=n2sinθ2nixi=nfxff=inisincenf1 for air and the apparent distanced=f. Thusd=2n2+1n1=2.4 cm1.4+4.6 cm1.33= (1.71429 cm) + (3.45865 cm)=5.17293 cm.The coin appears to be closer to the surfaceby(4.6 cm) + (2.4 cm)(5.17293 cm)= 1.82707 cm.Alternate Solution:

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Term
Fall
Professor
HOFFMAN
Tags
Snell s Law, Correct Answer, Total internal reflection, Geometrical optics