The problem statement, all variables and given/known dataMastering physics problem 23.76 difficulty rating: two bars=difficultA 2.0-cm-tall object is placed in front of a mirror. A 1.0-cm-tall upright image is formed behind the mirror, 180cm from the object.What is the focal length of the mirror?2. Relevant equationsh i / d i = h o / d oThin lens equation1/o + 1/i = 1/f3. The attempt at a solutionI made a previous attempt but I know why that was wrong. However, if I can see that the object is 2.0cm high and then its image is 1.0cm high. Does that imply that the distance of the object to the mirror is 180cm times 2 = 360cm so its image is 540cm from the mirror. I understand I can use the height ratio to find the magnification.
How does a lens form an image? See how light rays are refracted by a lens. Watch how the image changes when you adjust the focal length of the lens, move the object, move the lens, or move the screen. Interested in the size of an image or object. The ratio of the size of an image to the size of the object is called the magnification; it is given by, where is the height of the image, and is the height of the object. The second equality allows you to find the size of the image (or object) with the information provided by the spherical mirror. Image size of a person in a mirror? INTRO: A guy named Joe, who is 1.6 meters tall, enters a room in which someone has placed a large convex mirror with radius of curvature R equal to 30 meters. The mirror has been cut in half, so that the axis of the mirror is at ground level.
I just cannot see how I can find the obj distance or image distance. I still have 1/o + 1/i = 1/f and it is impossible to solve this equation when I still have three unkowns. Wow, thanks for the figure. My online homework still said that the answer of 160cm was still the wrong answer. I believe it is because we need to consider the thin lens equation after finding the obj and img distance. I cannot see how you are getting the focal length to be 160cm?
Image Size In A Mirror Mastering Physics 2
I'd be happy if you would just be able to help me get the object distance or image distance to the mirror. I'm not sure if I can see why you set 180cm=3x. Is that because of the three unknowns? That is the wrong way to go about doing it if so. Ok, answer is 40cm.x=dist from mirror to obj.x'=dist from mirror to image.m=magnificationL=vergence before reflectionL'=vergence after reflectionF=power of the mirrory=object heighty'=image heightf=primary focal pointF=-1/f or f=-1/FF+L=L'm=y'/y & x'/xL=1/xL'=1/x'abs(x)+abs(x')=180cmy'/y=1/2=m=x'/xso, x'/x = 1/2abs(x)=abs (2x') (substitution from equation above)abs(2x')+abs(x')=180cmabs(3x')=180cmabs(x')=60cmso.abs(x)=120cm (180-60)ok the problem said image was on opposite side of the mirror from the obj, so that would make x' a negative number. In mirrors, real images are on the same side of the object. In this case the image is virtual.so.x=120cmx'=-60cmL=1/1.2m=.83333 D (diopters)L'=1/.6m=-1.66666 DF=-2.5 D (L'-L)f=-1/-2.5=.4m or 40 cmfun problem!B.
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Woolverton, O.D.