Shrunken test 3

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Apr 3, 2024

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CH25%-A ray of light is reflected from 2 plane mirror surfaces as shown. What are the correct values of a and ﬂ" Value of 026° Value of 3 64°, 0 of reflect=to 0 of ingclinat, alpha = 26° & from the geom coordinated & A rules, sum of all 6 of a A is 180° for the right angled A, 90 + -alpha =180 alpha + § = 90, B = 90 - alpha = 64° ¥-Two mirrors are arranged as shown in the drawing. Light is incident from the right on the first mirror at an angle of 70°. The light reflects toward the second mirror and is reflected at angle €. Determine the angle 6. law of reflection states that the angle of incident = the angle of reflection, BP= normal to the mirror 1, QC= normal to the mirror 2, From the fig, mZBCO=m«£QCD=700, mgZBCO=900—m<£QCB=900—700=20°, m¢BCO=200, Now from 4BOE+mgBOC=1800, 450+m¢ BOC=1800, mgBOC=1350, From A law mg£QBC=250, Now m£PBQ=900, As PB = normal to the mirror 1, mgZPBC= Angle of reflection for the mirror 1, Now m£PBC=900—250=650 W-A concave mirror has a radius of curvature of 60.0 cm. How close to the mirror should an object be placed so that the rays travel parallel to each other after reflection? f=r/2=>f=60/2=30cm W-A concave mirror in an amusement park has a radius of curvature of 4.0 m. A child stands in front of the mirror, she appears 3.0 times taller than her actual hgight,If.the imag is upright, how far is she standing from the mirror? f=r/2=>1=4.0/2=2m, expresion for magnification=m= - di/do=>di= - m*do=>di= - 3.0(do), then, 1/£=1/di+1/do, subs 2.0m for f, and -3do for di, (1/2.0 W-An object is 1.0 m in front of a mirror. A virtual image is formed 10.0 m behind the mirror. What is the radius of curvature of mirror? 1/f=1/di+1/do and R=2f, 1/f=1/10m+1/1m, 1/f= -9/10m =>f=10/9m= 1.1m W-A spherical concave mirror has a radius of curvature of 6.0 cm. At what distance from the mirror should a6.0- cm object be placed to obtain an image that is 48 cm tall?1/f=1/di+1/do, focal length of mirror=f, distance of image=di, object distance=do. Rad cury=R=6.0cm, Object hight=H0=6.0cm, Image height=Hi=-48cm (-) cuz same side as object, f=R/2=>f=6cm/2=3cm, 1/do=1/f-1/di=>1/do=1/3c¢m-1/48cm=>1/d0o=17/48=> do=48/15=3.2cm=~3.4cm W-A convex mirror has a radius of curvature of 0.50 m. Where must an object be placed in front of the mirror such that the image is formed 0.15 m behind the mirror?1/do-1/0.15= - 2/0.5=>d0=0.375~0.38m in front of the mirror. W-A concave mirror is found to focus parallel rays at a distance of 18 cm. Where is the image formed when an object is placed 6.0 cm in front of the mirror?f>0 cuz mirror converges rays. do=6cm, (1/ di)= (1/f)-(1/do)=>(1/ di)= 1-3)/18= - 2/19=>1/di= - 1/9=>di= - 9cm (-) means it is behind the mirror=9.0 cm behind the mirror ¥-A 0.127-m pencil is oriented perpendicular to the principal axis of a concave spherical mirror that has a radius of curvature of 0.300 m. What are the image distance and the image height if the pencil is 0.250 m from the mirror? Radius of SHLY. of a concave spherical mirror (R)=0.300m, focal length of mirror f=R/2=>f=0.300/2=0.15m, do=0.25m, ho=0. 127m mirror equatlon 1/ di)= (1/9)-(1/do)=>(1/ d;)=(f*do)/do-f=>di=0.15*0.250/0.250-0.15=0.375m i/do)*ho=>hi=( - 0.375/0.250)*0.127m=>h’= - 0.1905m= - 0.191m ¥-The inverted image of a light bulb is formed on a screen located 9.00 m from a spherical mirror. The image is 4 times larger than the light bulb,Determ. the obj distan & type of mirror used. B= - di/do=hi/ho=>-4=-di/do => do=9/4 =2.25m, mirror formla is: 1/f=1/di+1/do=>f=(di*do)/di+d0=9%2.25/9+2.25=1.8m>0 +2.25 m, concave=>0 W-A convex mirror with =58 cm is used to form an image that is 29cm behind the mirror, What is obj dist? f= - 58cm, di= - 29cm,do=di*f/di-f=>do=(-29*-58)/-29- -58=58c m=>d0=58cm, B=di/d0=29/58=0.5cm +upright +58 cm ¥-An object with a height of 3.0-cm is placed 8.0 cm in front of a mirror. The virtual image is 4.0 cm further from the mirror when the mirror is concave than when it is planar. (a) Determine the focal length of the concave mirror. For normal mirror do=di=8, for concave do=8 and di=12_1/f=1/do+1/di=>1/f=1/8-1/12=24cm, (b) Determine the image height in the concave mirror. ho=3, hi=ho*(di/ hi=3*12/8=4.5cm CH26 . ¥-When certain light rays pass from a vacuum into a block of an unknown material, the measured index of refraction of the material is 3.00. What is the speed of light inside the block? index of refract n=3.0, speed of light inside the block v=c/n=>v=(3*10*m/s)/3=_1.0 x 10® m/s ‘I‘- The speed of light in materlal Ais 14 tlmes as large as it is in material B. What is the ratio of the refractive 4 What is the value of the distance D? use Snell's law= nq*sin(0i) = n; sm(er). n; = refractive index air = 1 i=50degrees n:=ref index glass = 1.59 Then, 1*sin(50) = 1.59*sin(r), sin(r) = sin(50)/1.59 = 0.481, r = asjn(0.481) = 28.8° ooem is 0 of refraction B) from the figure tan(r) = D/4. D = 4*tan(r) = 4*tan(28.8) = 2.2 cm ¥-A fish swims 2.00 m below the surface of a pond. At what apparent depth does the fish appear to swim if viewed from directly above? The index of refraction of water is 1.33, ldwater= real depth/ apparent depth, apparent depth = realdenthluwaer =2.00/1.33=1.50m ¥-A scuba diver shines a flashlight from beneath the surface of water (n = 1.33) such that the light strikes the water-air boundary with an angle of ingide of 37°. At what angle is the beam refracted?n*r=1,we know from snell's, law: n; »sin 6i)/ n2*(sin Or) => Or=sin-'(1.33*sin(37°))=53.17~53° ¥-Light with a wavelength of 589 nm in a vacuum strikes the surface of an unknown liquid at an angle of 31.2° with respect to the normal to the surface. If the light travels at a speed of 1.97 x 10® m/s through the liquid, what is the angle of refraction? A=589nm=589x10-m Refractive index of air n1=1, incident angle 61=31.2°, sp of light V1=3x108m/sec, gp light in liquid V2=1.97x108m/sec, 6 of refraction?n+*sin(incidence®) = n2*sin(refraction®), n=refraction index, gnell's. law: ni/nz2=sin(01)/sin(02)=V1/V2 , n1/n2=V1/Vz, n2=(V1/V2)xn1, n2=(3x108/1.97x108)x1, n2=1.522, Snell's law, nisin(61)=n2sin(62)=> sin(62)=(1/1.522)*sin(31.2°), Now substitute 62=19.89° => 62=19.9° W-A light ray is traveling in a diamond (¢ = 2.419). If the ray approaches the diamond-air interface, what is the minimum angle of incidence that will result if all of the light is reflected back into the diamond? The index of refraction for air is 1.000._N9*sin(0j)=n2a*sin(0r)=> sin(6j)=na/nd*sin(0r),sin(6j)=(nair/ndiamond)*sin(90)=> Bi=sin-1(1/2.419)*1, Bi=sin-1(0.41339), 6=24.418°=>24.42° P-Light propagates from soda lime glass (n2 = 1.518) into Pyrex glass (n/ = 1.473). Determine the critical angle for this situation. N9*sin(0j)=n2"sin(6c), 6c=critical angle, 6c can be calculated w/Snell's law setting the refraction angle to 90°. N1*sin(90)=nz"*sin(6¢c)=> N1*1=nz"sin(6c)=>0c=sin""(n1/nz)=>6c=sin"'(1.473/1.518)=76.01° ¥-A ray of light originating in oil (» = 1.21) is incident at the Brewster angle upon a flat surface of a quartz crystal (n = 1.458). Determine the angle of incidence for this ray. tanBz=n,/n;=> tanOs=1.458/1.21, Bs= tan"!(1.20495)=50.3°~50° ¥-What is the Brewster angle when light is reflected from a plastic plate (» = 1.575) submerged in ethyl alcohol (n = 1.362)? tan@g=ny/n;=>tanBz=1.575/1.362, Op= tan"! (1.575/1.362)=49.14° W-An object is placed 6.5 cm from a thin converging lens with a focal length of 14 cm. Which one of the following statements is true concerning the image? Focal L=14cm, u=-6.5, (1/v)-(1/u)=1/f=>(1/v)=(-1/6.6)=1/14=>v=-12.13¢cm , m=v/u=>m=-12.13/(- 6.5)=1.866=>virtual image. The image is virtual and 12 cm from the lens ¥-A converging lens is used to focus light from a small bulb onto a book. The lens has a focal length of 10.0 cm and is located 40.0 cm from the book. Determine the distance from the lens to the light bulb. 1/f=1/p+1/g_, f=+10cm (+cuz lens is converging), g=+40cm( + means it is a real image), 1/10=1/p+1/40=>p=13.3cm W-An object with a height of 4.0 cm is placed 30.0 cm from a lens. The resulting invert jmag has a height of 1.5 cm. What is the focal length of the lens”h1=4cm, u=-30cm, image height is h2=-1.5¢cm(-conversing lens), M=h2/h1=v/u=>v=-30(-1.5/4)=11.25cm, lens formula=1/f=1/v-1/u=>1/f=1/11.25-1/- 30=>1/f=3.6666666/30=>f=30/3.6666666=8.182cm=8.2cm ¥-A 2.00-cm tall object is placed 40.0 cm from a lens. The resulting image is 8.00-cm tall and upright relative to the object. Determine the focal length of the lens,hy=2cm, u=40cm, hi=8cm The focal length is said to be the measurement of the strength of the convergence and the divergence of the lens. -v/u=hi/ho=> -v=(8/2)*40=- 160cm, 1/f=1/v+1/u, 1/f= -1/160+1/40=>f=53.3cm (+ sign means lens is convex) ¥-A diverging lens has a focal length of 15 cm. A 5-cm object is placed 35 cm from the lens. f=--15¢m, u=-35¢cm_ . object length=5cm, (a) Determine the approximate distance between the object and the image. 24 cm V/f=1/v-1/u=>1/--15=1/v+1/35=>v=(35x15)/(-35-15)=-10.5cm, dist bet image & object d=v-u=>d=-10.5-(-35)=24.5cm

(b) What is the magnification of the image? +0.3 m=v/u=>m=-10.5/-35=0.3 (¢) Which terms most accurately describes image? Sign of v is (-) so it’s virtual, magnitude is + so it is upright. ¥-A 4.0-cm object is placed 30.0 cm from a converging lens Lens that has a focal length of 10.0 cm as shown in the diagram. Objecﬂ . é Note: The diagram is not drawn to scale. (a) Where is the image located? lens 1/f=1/v-1/u, 1/10=1/v-(1/-30)=15 cm =>v=15cm to the right of the lens (b) Determine the height and orientation of the image. Magnification m=hi/ho=v/u>m=-15/30=-0.5=>mag is (-) so it is inverted . . and smaller than obj size 4*0.5=2 cm and inverted —— 30.0 cm —» 110.0 cm! - (c) A second converging lens is placed 20.0 cm to the right of the lens shown in the figure. Determine the focal length of the second lens if an inverted image (relative to the object in the diagram) is formed 13.3 cm to the right of the first lens. u=-5 since image of first lens acts as an object for second lens, v=-6 7cm since given final image is formed at 13.3cm to the right of first | 1/f= 7 1/-5)=>f=335/17=19.7 ¥-Two converging lenses, each with a focal length of 0.080 m, are used in combmatlon to form an image of an object that is located 0.24 m to the left of the left lens in the pair. The distance between the lenses is 0.16 m. £i=£,=0.080m. obj dist to lens X;= - 0.24m( (-) cuz obj is to left of lens), d between the lenses=0.16m (a)Where is the final imag located relative to the lens on the right?Lens frmla: (1/ di)-(1/ do)=(1/f), £=0.080m, d,=- 0.24, di=( d, *f)/(d, +)=( -0.24 *0.08)/(-.24 +0.08)= (-0.0192/-0.16)=0.12m =>from 2" lens obj dist d’¢=0.16- 0.12=0.04m, d’;=( d’o *NH/(d2o+)=( -0.04 *0.08)/(-0.04 +0.08)= (-3.2*10-3/0.04)= -0.08m to the left of lens (b) What is the total magnification of this lens combination? Mag frmla B= (di/do)=>(.12/-.24)*(-0.08/-0.04)=- ¥-A 1.5-cm object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens. (a) What is the location of the final image with respect to the object? (1/ d;)-(1/ do)=(1/f), do= -0.5m, f= -.20m, for conyerg lens £=0.17m, diverg lens di=> (1/ di)=( 1/f)-(1/ do)=> (1/d;)=(1/-0.20)-(1/-0.50)=1/7= -0.143m, do= - (0.143 + 0.08)= - 0.223m, so, (1/di)-(1/do)=(1/H)=>(1/d;)-(1/-.223)=(1/. 17)—>(1/d|)—(1/ 17)+(1/-. 2230)—>d1—0 715m, now B,= (di/do)=( 0. 715m/-0 223)= - 3.2063. the final image location is 715 nght to the converging lens, so, its dlstance 1 d 1. Total B=p1*p2=> B=0 286*(—3 M.); -0.9170, Hi=p*H, = -0.9170%0.015= -0.01375m=> - 1.375cm=1.4cm mverted (b) height and orientation with respect to the original object of the final image? 1.4cm, inverted (c) If the height of theobiectis 1.0 cm,whatisthe height of image? -0.9170*0.010=-0.00917m=> - 0.917cm = 1.2 cm (d) Which pair of terms most accurately describes the final image? real, inverted ¥-A stamp collector used a magnifying glass with a focal length of 24.0 cm to observe his favorite stamp. The image of the stamp was located at his near point distance of 30.0 cm. What was the approximate angular magnification of magnifying glass when the collector observed his stamp? Equation for B=1+(near point/f)=> 3=1+30/24=2.25 CH27 ¥-A double slit is illuminated with monochromatic light of wavelength 6.11 x 10>nm. The m =0 and m = 1 bright fringes are separated by 1.8 cm on a screen which is located 3.5 m from the slits. What is the separation between the slits? B=AD/d=>d=AD/B=(611*10"m*3.5m)/1.8*10?m=1.188 x 10*m 1.2 x 10*m ¥-Two slits separated by 2.00 x 10~ m are illuminated by light of wavelength 625 nm.If the screen is 6.00m from the slits, what is the dist bet the m=0&m=1 bright fringe? B=AD/d-0=>B=((625*10-"m)(6m))/2*105 =0.1875m=>18.8cm ¥-In a Young's double slit experiment, the separation between the slits is 1.20 x 10~ m; and the screen is located 3.50 m from the slits. The distance between the central bright fringe and the second-order bright fringe is 0.0375 m. What is the wavelength of the light used in this experiment? B=y/2=0.0375m/2=0.01875, B=Fringe-width=(Distance 2 screen*))/(d=distbet slits). B=AD/d=>A=B*d/D=>A=((0.0185m)(1.2*10-*m))/(3.5m)=642.857*10m=643nm=642nm ¥-A 4.0 x 10*-nm thick film of kerosene (n = 1.2) is floating on water. White light is normally incident on the film. What is the visible wavelength in air that has a maximum intensity after the light is reflected? Note: the visible wavelength range is 380 nm to 750 nm. Thickness of film (t)=4.0*10’nm, n-1.2, x=n*ttn*t=2n*t, express of A=(2n*t)/m, here m is the order, m=0, A=(2*1.2*4.0*10? )/0+.5=1,920nm(not visible), then, m=1, A=(2*1.2*%4.0*10%) /1+.5=640nm, then m=2, A=(2*1.2*4.0*10?)/2+.5=380nm, then m=3, A=(2*1.2*4.0*10? )/3+.5=274.3nm, visible wavelength in the air that has a max intensity after the light is reflected is 380 nm P-A portion of a soap bubble appears yellow (A = 588.0 nm in vacuum) when viewed at normal incidence in white light. Determine the two smallest, non-zero thicknesses for the soap film if its index of refraction is 1.40. Equation for interference of light by soap bubbles: 2u*t=(m+1/2)A*(1), p=refractive index of soap film, t=thickness of soap film, m=order of interference & A=A or the light, for m=0, which gives the least possible thickness of film. 2*1 4*1—(0+1/2)588nm*( 10 “m/1nm)*( 1), 2.8*t=2.94*10""m=>t=1.05*10-7m=>105nm=smallest thickness of soap film 2*1.4*t=(1+1/2)588nm* (10 m/1pm)*(1). 2.8*t=8.82*10'm=>t=3.15*10-7m=>315nm=2"¢ smallest t of soap film W-A transparent film (n = 1.4) is depos1ted on a glass lens (n = 1.5) to form a non-reflective coating. What thickness would prevent reflection of light with wavelength 5.00 x 10? nm in air?n1=1.4, n2=1.5, A=500nm, the A inversely varies with the refractive index of the medium. t=A/4n1=>t=500/(4x1.4)=89.28nm=~89nm W-A lens that has an index of refraction of 1.61 is coated with a non-reflective coating that has an index of refraction of 1.45. Determine the minimum thickness for the film if it is to be non-reflecting for light of wavelength 5.60 x 10? nm. t=A/4n1=>t= (560 x 10 nm)/(4*1.45)=9.655x10*m~9.66x10*m ¥- Light of wavelength 625 nm passes through a single slit of width 0.320 mm and forms a diffraction pattern on a flat screen located 8.00 m away. Determine the distance between the middle of the central bright fringe and the first dark fnnge k—625nm—625x10 ’m, w=0. 320mm—320x10 ‘m, D—8m mk—w*sm(e), angle is given by . /8 =156.2*10“*m=~1.56cm ¥-Light with a wavelength of 600.0 nm is incident on a single slit of width 6.5 pm. The resulting diffraction pattern is observed on a nearby screen and has a central maximum of width 3.5 m. What is the distance between the screen and the slit? Y=A*D/d, y=(600*10-m*D)/6.5*%10°m, the central bright fringe should be double to this distance y=3.5/2, y=1.75m, 1.75=600x10-9xD/6.5*10°m=>D=18.9583m~19m ¥-A monochromatic beam of microwaves with a wavelength of 0.052 m is directed at a rectangular opening of width 0.35 m. The resulting diffraction pattern is measured along a wall 8.0 m from the opening. What is the distance between the first- and second-order dark fringes? A=0.052m, d=0.35m, condition for dark fringe by a single slit is d*sin(@)=nA, for first order dark fringe is d*sin(0;)=nA, sin(01)=0.052/0.35=0.148, 0;=sin"'(0.148)=8.54°, the distance of first fringe from the center is y1=Ltan(01)=8tan(8.54°)=1.20m, for second order dark fringe is d*sin(01)=2*A, or s1n(62)—2Md—2(0 052/0 35) =0.297, 0,=sin"1(0.297)=17.27°, the distance of second fi'mge from the center is 0 .27°)=2.48 h ¥- Light with a wavelength of 644 nm uniformly illuminates a single sllt What is the width of the sllt 1f the first- order dark fringe is logat at A_= 0.125°? A=644nm=644x10"m, 6=0.125°, d*sin(0)=nA, d*sin(0.125°)=1*644*10°m, d =1*644*10"°m/sin(0.125°), d=2.95*10"*m W-When light with a wavelength of 425 nm uniformly illuminates a single slit, the central bright fringe, observed on a screen located 0.630 m from the slit, has a width of 0.0166 m. What is the width of the slit? A=425nm=>425x10"m, the width of the slit, y=0.0166m, d from screen, D=0.63m, d=AD/y, for the single slit at the central bright maxima the width Y=y/2, Y=0.0166/2=>Y=0.0083m, the single slit diffraction for the central bright fringes the width will become half, d=(425*10-*0.63)/0.0083=3.23*10"m ¥-The Hubble Space Telescope in orbit above the Earth has a 2.4 m circular aperture. The telescope has equipment for detecting ultrav1olet hght What is the minimum angular separation between two objects that the Hubble ¥-The headlights of a car are 1.6 m apart and produce light of wavelength 575 nm in vacuum. The pupll of the eye of the observer has a diameter of 4.0 mm and a refractive index of 1.4. What is the maximum distance from the observer that the two headlights can be distinguished? Distance bgfyy, headlights D=1.6m, A=575nm=>575x10"m, Diameter of the eyes is d=4*10m, n=1.4, A in medium A’=A/n=>1=410.71nm=>410.71x10"m, x=1,22MD/d)=>D=x*d/1.22), D=12.77x10’m=~13km W-Two stars are just barely resolved by a telescope with a lens diameter of 0.500 m. Determine the angular separation of the two stars. Assume incident light of wavelength 500.0 nm. According to the Raleigh criteria for resolving power, the angular separation between 2 stars is: 6=1.220/D, 8=1.22*500 x10-°m/0.500m=1.22*10-6rad ¥-Two candles are lit and separated by 0.10 m. If the diameter of the pupil of an observer's eye is 3.5 mm, what is _ maximum distance that the candles can b away from the ghsery, & b seen as 2 light sources? Use 545nm for A of light in the eye. G=1.22*lamda/D. also G=y/R=>y/R=1.22*lamda/D=>0.1/R=(1.22*545*10?)/(3.5*10-*)=>R=526m=~530m ¥-Red light of wavelength 600.0 nm is incident on a grating. If the separation between the slits is 5.0 x 10 m, at what angle does the first principal maximum occur? d*sin(6)=m*A=>sin()=1*600*10m/5.0 x 105 m=1.2*10"2rad W-A diffraction grating that has 4500 lines/cm is illuminated by light that has a single wavelength. If a second order maxi is ghsery, at 0 42° with respect to the central maximum, what is the waygl of this light2Eqrmula diffraction due to grating is: d*sin(6)=m*\, d=spacing betw adjacent lines in the grating, 9-where maximum is observed, mis order of the maximum, A=) of the light. D=1/4500lines/cm=2.22x10° m, A=(2.22x106

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