Concave mirrors - where is the object

  • Thread starter Thread starter danago
  • Start date Start date
  • Tags Tags
    Concave Mirrors
AI Thread Summary
A concave mirror with a focal length of 1m is used to magnify an object by a factor of 4, leading to the calculation of the object's position relative to the mirror. The initial calculations incorrectly used the magnification formula without accounting for the negative sign in the relationship between object distance and image distance. The correct approach involves using the mirror formula and recognizing that the magnification is negative, resulting in the object distance being 0.75m from the mirror. This conclusion aligns with the textbook answer, confirming the correct positioning of the object. Understanding the sign conventions in optics is crucial for accurate calculations.
danago
Gold Member
Messages
1,118
Reaction score
4
"A concave mirror of focal length 1m is used to magnify an object by a factor of 4. Where is the object in relation to the pole of the mirror?"

Heres what i did:
<br /> \begin{array}{c}<br /> M = \frac{v}{u} \\ <br /> \therefore 4 = \frac{v}{u} \\ <br /> \therefore v = 4u \\ <br /> \end{array}<br />

<br /> \begin{array}{c}<br /> \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ <br /> \therefore 1 = \frac{1}{u} + \frac{1}{{4u}} = \frac{5}{{4u}} \\ <br /> \therefore u = 1.25m \\ <br /> \end{array}<br />

Where am i going wrong? According to my textbook, the answer is 0.75m.

Thanks in advance,
Dan.
 
Physics news on Phys.org
You are forgeting that M = -di/do

Formula = (1/do + 1/di) = (1/f)

So 4do = - di

1/do - 1/4do = 1

1/4do - 4/4do = 1

3/4do = 1

do = .75
 
Last edited:
Alt+F4 said:
3/4do = 1

do = .75

Right.

3/(4do) = 1 so do = .75
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top