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The focal length of a mirror. Help me maybe? 
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#1
Nov1008, 05:40 PM

P: 24

1. The problem statement, all variables and given/known data
Mastering physics problem 23.76 difficulty rating: two bars=difficult A 2.0cmtall object is placed in front of a mirror. A 1.0cmtall upright image is formed behind the mirror, 180cm from the object. What is the focal length of the mirror? 2. Relevant equations h_{i} / d_{i} = h_{o} / d_{o} Thin lens equation 1/o + 1/i = 1/f 3. The attempt at a solution I 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. 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. 


#2
Nov1108, 03:15 AM

P: 8

Hey
check this: ok so you focal distance is going to be 160cm because we can say that the 180=3x and the ratio between the sizes of the two objects is 2:1 


#3
Nov1108, 01:02 PM

P: 24

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? 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.



#4
Dec909, 07:45 PM

P: 1

The focal length of a mirror. Help me maybe?
ok, answer is 40cm.
x=dist from mirror to obj. x'=dist from mirror to image. m=magnification L=vergence before reflection L'=vergence after reflection F=power of the mirror y=object height y'=image height f=primary focal point F=1/f or f=1/F F+L=L' m=y'/y & x'/x L=1/x L'=1/x' abs(x)+abs(x')=180cm y'/y=1/2=m=x'/x so, x'/x = 1/2 abs(x)=abs (2x') (substitution from equation above) abs(2x')+abs(x')=180cm abs(3x')=180cm abs(x')=60cm so...abs(x)=120cm (18060) 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=120cm x'=60cm L=1/1.2m=.83333 D (diopters) L'=1/.6m=1.66666 D F=2.5 D (L'L) f=1/2.5=.4m or 40 cm fun problem! B. Woolverton, O.D. to be 


#5
Sep2210, 03:52 PM

P: 2

40cm isn't right either, unfortunately. And I don't have the answer just yet! Hopefully, I'll have a solution by tomorrow!



#6
Sep2310, 12:47 AM

HW Helper
Thanks
P: 10,634

An upward image means that the image is virtual. If it is a concave mirror, the virtual image is bigger than the object. So the mirror in the problem is convex. Therefore the focal length is negative, the image distance is negative. i+o=180, i/o=0.5, that is o=120 and i=60. Plug in to the lens/mirror equation.
ehild 


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