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## MCAT

### Course: MCAT > Unit 8

Lesson 19: Thin lenses- Thin lenses questions
- Thin lens sign conventions
- Convex lenses
- Convex lens examples
- Concave lenses
- Object image and focal distance relationship (proof of formula)
- Object image height and distance relationship
- Thin lens equation and problem solving
- Multiple lens systems
- Diopters, Aberration, and the Human Eye

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# Object image height and distance relationship

Object Image Height and Distance Relationship. Created by Sal Khan.

## Want to join the conversation?

- How do you solve problems that deal with normal adjustments and angular magnifications of both microscopes and telescopes? Could you give me an example? thanks!(21 votes)
- Considering the fact that the ratios of di/do and hi/ho are equal to the magnification of the image, I would say this is reasonably related.(8 votes)

- In terms of Magnification, doesn't hi/ho = -di/do, the negative sign matters no?(8 votes)
- Yes (the next video talks about the positive and negative conventions)(1 vote)

- Where can i find problems for optics online or in which books?(3 votes)
- you can refer the following links

http://www.sparknotes.com/physics/optics/geom/problems_2.html

http://www.physicstutorials.org/home/exams/optics-exams-and-solutions/optics-exam1-and-solutions

http://academics.wellesley.edu/Physics/phyllisflemingphysics/108_p_geooptics.html

or refer the following powerpoint presentations

http://www.fisica.net/optica/optics_textbook.pdf

http://optics.byu.edu/BYUOpticsBook_2011c.pdf

otherwise any proper phusics college guide will have numericals based on the above concept(5 votes)

- the correct lens formula is 1/f = 1/v - 1/u , isn't it or 1/f = 1/di - 1/do(1 vote)
- See, what u r talking about is called 'sign convention'......where u assume a Cartesian plane in which the optical centre of the lens acts as the origin and the principal axis as the x-axis......so any distance on the left side of lens is taken as negative (u= do= distance of the object from the lens) and any distance on the right is taken as positive(v= di= distance of image from the lens)......also, when the image is inverted the height of the image is taken as negative for the same reason....(5 votes)

- How can I calculate the object or image height if I have the focal length, object distance and image distance? Thank you :)(3 votes)
- where is the first video for this video?(2 votes)
- Does this formula apply to parabolic convex and concave mirrors and to convex and concave lenses?(1 vote)
- Hi,

This wasn't very clarified in the video so can someone plz help me out:

Say you have an image of 30 cm tall flower and it is captured by a camera with a 28mm lens. If the image formed behind the lens is 29mm behind the lens, how far Is the lens from the rose? how high is the image?(1 vote)- Our key formula is 1/do + 1/di = 1/f

In your question,

f = focal distance = 28 mm

di = image distance = 29 mm

do = object distance = what we want to find

So, that gives us:

1/do + 1/29 = 1/28

1/do = 1/28 - 1/29

1/do = 1/ 812

do = 812 mm (has to be mm since 28 and 29 were mm!)

do = 81.2 cm

Then, for the height of the image, we use the formula Sal finds in this video:

hi/ho = di/do

hi/300mm = 29mm/812mm (change EVERYTHING to mm to be safe!)

hi = (29/812)*(300mm)

hi = 10.7mm(1 vote)

- Does this mean that you could also write the equation found in the previous video as (1/f)=(1/ho)+(1/hi) ? Also, can this formula also somehow be used to find the focus of a convex/concave parabolic mirror?(1 vote)
- no,as only ho/hi=u/v for a lens .it doesn't implies that ho=u & hi=v.(1 vote)

- Can anyone explain me what is angular magnification?(1 vote)

## Video transcript

In the last video, I showed
that a bunch of triangles are similar to each
other to come up with the relationship
between the focal length, the distance of an object
from the convex lens, and the distance of the image
of an object from that convex lens. And I realized that there was
one extra low-hanging fruit based on all of the
geometry that I had done, another relationship that
might come in useful. And that's the relationship
between the size of the object-- or maybe
since it's an arrow here, we can call it the
height of the object, and the height of the image. And we really set up
everything already. We already figured out that
this triangle over here is similar to this
triangle down here. And we figured out
in the last video that this triangle over here
is similar to this triangle over here. And since these
two are similar, we could say that A is to
B-- we did this over here. I'll rewrite it. A is to B as-- and both
of those are the sides opposite the right angles of
these two similar triangles. So that's going to be the same
thing as the ratio of the sides opposite this yellow
angle right over here. So in this triangle over here,
since we started with A first, it's this height
right over here. Now what is this
height right over here? This is the height
of the object. So this is the
height of the object is to-- Now what is
this opposite side of this yellow angle
right over here? Well, this is the
height of the image. Or we know from the
last video the distance of the object to the
distance of the image is the same thing as A to B. So this is going to be
the same thing as this. So the ratio of the
distances is also the same thing as the
ratio of their heights. So let me write it this way. So the ratio of the distance
from the object to the lens, to the distance from
the image to the lens, is the same as the
ratio of the height of the object to the
height of an image, or to the image of that object. So I just wanted to do that
little low-hanging fruit there, since we set up all of
the mechanics already. Anyway, hopefully you
found that useful.