# Optical Size versus Actual Size

#### Libertine

##### New member
Hi all,

I just joined this Forum to get an answer to a question really, it's not for Collage or Uni or anything just a curiosity really.
Mathematics isn't one of my strong points and I hope this is in the right section.

What I'd like to know is.... If an Object is 3 miles away, and this is what complicates things a little (for me at least), it is 3mm wide by 4mm heigh through a pair of Binoculars (15x70).
What size would the Object be?

I'm also not 100% sure of the distance so if anyone could help me out with this could you also do the same thing but with the Object at 2 miles and also 4 miles away if possible?

If anyone can help me with this I'd be really grateful.

#### CaptainBlack

##### Well-known member
Re: Help with a question please.

Hi all,

I just joined this Forum to get an answer to a question really, it's not for Collage or Uni or anything just a curiosity really.
Mathematics isn't one of my strong points and I hope this is in the right section.

What I'd like to know is.... If an Object is 3 miles away, and this is what complicates things a little (for me at least), it is 3mm wide by 4mm heigh through a pair of Binoculars (15x70).
What size would the Object be?

I'm also not 100% sure of the distance so if anyone could help me out with this could you also do the same thing but with the Object at 2 miles and also 4 miles away if possible?

If anyone can help me with this I'd be really grateful.
$$x$$mm high through a pair of binoculars means nothing, if you mean the same apparent size as a $$x$$mm high object held at arms length then that does mean something.

$$15$$ times magnification means that the apparent angle subtended by the target is $$15$$ times greater than without the glasses. So if the real size of the object is $$X$$ at a range of $$R$$ (all in metres) the true angular size is $$X/R$$ radian or $$(180/\pi)\times (X/R)$$ degrees. So the apparent size through the glasses is $$15\times (X/R)$$ radian which is the same apparent size as an object $$15\times (X/R) \times 0.75$$ metres at arms length ($$0.75$$ metres approximately).

CB

#### Libertine

##### New member
Re: Help with a question please.

Hi CaptainBlack.

Sorry, I meant that through the Binoculars the Object was 3mm in Width and 4mm tall.

#### CaptainBlack

##### Well-known member
Re: Help with a question please.

Hi CaptainBlack.

Sorry, I meant that through the Binoculars the Object was 3mm in Width and 4mm tall.
Please read my earlier post for an explanation about why that is either incomplete information or just nonsense.

Or to put it another way; how did you measure its height and widths as seen through the glasses?

CB

#### HallsofIvy

##### Well-known member
MHB Math Helper
Re: Help with a question please.

Hi all,

I just joined this Forum to get an answer to a question really, it's not for Collage or Uni or anything just a curiosity really.
Mathematics isn't one of my strong points and I hope this is in the right section.

What I'd like to know is.... If an Object is 3 miles away, and this is what complicates things a little (for me at least), it is 3mm wide by 4mm heigh through a pair of Binoculars (15x70).
What size would the Object be?
"15x70" means that each object lens is 70 mm (7 cm) in diameter and the focal length is 15 times the diameter or 1050 mm (105 cm). Neither tells you what the magnification is. The magnification is th focal length of the objective lens divided by the focal length of the eyepiece- which is not given here.

I'm also not 100% sure of the distance so if anyone could help me out with this could you also do the same thing but with the Object at 2 miles and also 4 miles away if possible?

If anyone can help me with this I'd be really grateful.
[/QUOTE]

#### CaptainBlack

##### Well-known member
Re: Help with a question please.

"15x70" means that each object lens is 70 mm (7 cm) in diameter and the focal length is 15 times the diameter or 1050 mm (105 cm).
I'm afraid you are wrong, the first figure is the magnification - linear or angular, they are the same thing (the second as you say is the diameter of the objectives in mm).

You can look this up on almost any site that talks about bins, the Wikipedia article for example.

CB

#### Libertine

##### New member
Re: Help with a question please.

HallsofIvy.

CaptainBlack is right, the 15x is the magnification, the 70 is the aperture.

Hi CaptainBlack,

Ah I see now, lol if you hadn't already guessed Maths isn't my strong point.
I also now think I missed judged the distance by quite a way, I didn't take into account how high up the object was.

I think this will now make sense.

If object A is 6 miles away and is the same size (through 15x Bins) as object B which is 3mm wide by 4mm tall (with the naked eye) held at arms length.
What size is object A?

#### CaptainBlack

##### Well-known member
Re: Help with a question please.

HallsofIvy.

CaptainBlack is right, the 15x is the magnification, the 70 is the aperture.

Hi CaptainBlack,

Ah I see now, lol if you hadn't already guessed Maths isn't my strong point.
I also now think I missed judged the distance by quite a way, I didn't take into account how high up the object was.

I think this will now make sense.

If object A is 6 miles away and is the same size (through 15x Bins) as object B which is 3mm wide by 4mm tall (with the naked eye) held at arms length.
What size is object A?

Assume that arm's length is 750mm, 1 mm at arms length subtends an angle of 1/750 radian, so an object that subtends 1/750 radian through the binoculars subtends (1/750)/15 radian without the binoculars.

6 miles is 9656 metres.

An object with a linear dimension of x metres at 9656 metres subtends x/9656 radian, if this is the same apparent size as 1 mm at arms length we have:

x/9656=(1/750)/15

or:

x=9656/(750*15)=0.858m

So every 1mm at arms length maps onto 0.858m at 6 miles, so the dimensions of your object at 6 miles are 3*0.858=2.74 metres and 4*0.858=3.43 metres

CB

#### Libertine

##### New member
Re: Help with a question please.

Brilliant!
Thanks mate, I really appreciate it.