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:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top