Optics (Lens and Mirrors) Questions

[MedicineForLife]

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Hi!
I am confused when to used the (-) negative sign in the magnification equation:

m = -i/o (where i and o are the distances of the image and object from the lens or mirror, respectively)

For example, in this problem:
An object is placed in front of a concave mirror having a radius of curvature of 0.3m. If you want to produce first a real image and then a virtual image 3x as high as the object's, find the object distance required in each case.


How would I use the m = -i/o? For example, for the real image, since real images are positive, do I set m into a negative value?

Hope my question is clear. Please help me. Thanks so much!

 

[TieuBachHo]

I am not sure if this is a homework or a MCAT problem but I'll guide you to work it out on your own (Most ppl here won't respond to HW problems).

1- Since this is a concave mirror, you set r = 0.3m. From here you can find the relationship between image (i) and object (o).

2- For simplicity, you can set m = - 3 for REAL image and vice versa.

==> from 1 and 2 you can easily solve this problem.

To answer your question, yes, you set m to a negative value for REAL image so that i will be positive.

 

[MedicineForLife]

TieuBachHo: Thank you so much for the time you took to help me out!

 

[BerkReviewTeach]

TieuBachHo: Thank you so much for the time you took to help me out!

As was mentioned in Tieu's post, homework help is frowned upon here, but given it's been a month, I assume it's safe to respond now. Plus, this is a great question that can easily be turned into an MCAT question. You can solve it using the thin lens equation.

Let's first consider a real image. f = 0.3 and di = 3do (the value for an inverted image that is three times the size of the object).

• Here goes the math:
  1/f = 1/do + 1/di
  1/0.3 = 1/do + 1/3do = 3/3do + 1/3do = 4/3do
  1/0.3 = 4/3do so inverting means 0.3 = 3do/4
  do = 0.3 x (4/3) = 0.4

do = 0.4, so di = 1.2 and m = -1.2/0.4 = -3 (inverted and 3x larger)

Let's now consider a virtual image, which results when the object is inside of f, meaning do is likely around 0.2 or so. f = 0.3 and di = -3do (the value for an upright image that is three times the size of the object).

• Here goes the math:
  1/f = 1/do + 1/di
  1/0.3 = 1/do + 1/-3do = 3/3do - 1/3do = 2/3do
  1/0.3 = 2/3do so inverting means 0.3 = 3do/2
  do = 0.3 x (2/3) = 0.2

do = 0.2, so di = -0.6 and m = -(-0.6)/0.2 = +3 (upright and 3x larger)

 

[GASPER20]

As was mentioned in Tieu's post, homework help is frowned upon here, but given it's been a month, I assume it's safe to respond now. Plus, this is a great question that can easily be turned into an MCAT question. You can solve it using the thin lens equation.

Let's first consider a real image. f = 0.3 and di = 3do (the value for an inverted image that is three times the size of the object).

• Here goes the math:
  1/f = 1/do + 1/di
  1/0.3 = 1/do + 1/3do = 3/3do + 1/3do = 4/3do
  1/0.3 = 4/3do so inverting means 0.3 = 3do/4
  do = 0.3 x (4/3) = 0.4

do = 0.4, so di = 1.2 and m = -1.2/0.4 = -3 (inverted and 3x larger)

Let's now consider a virtual image, which results when the object is inside of f, meaning do is likely around 0.2 or so. f = 0.3 and di = -3do (the value for an upright image that is three times the size of the object).

• Here goes the math:
  1/f = 1/do + 1/di
  1/0.3 = 1/do + 1/-3do = 3/3do - 1/3do = 2/3do
  1/0.3 = 2/3do so inverting means 0.3 = 3do/2
  do = 0.3 x (2/3) = 0.2

do = 0.2, so di = -0.6 and m = -(-0.6)/0.2 = +3 (upright and 3x larger)

i thought focal length= 1/2r ? so shouldn't it be 1/0.15=1/3do +1/do; do=0.2m

 

[BerkReviewTeach]

i thought focal length= 1/2r ? so shouldn't it be 1/0.15=1/3do +1/do; do=0.2m

OOOPPPSSS. You are 100% correct. I misread the question as f = 0.3 and not R = 0.3. Careless errors are a killer. Your answer is correct.