Does the size of a projected image double if you half the focal length of the lens?

[68degrees]

I have a friend who has a projector with a 300mm lens. The projected image on the screen is 3 feet by 3 feet. (this is hypothetical). If he
changes out the condenser and replaces the 300mm lens with a 150mm lens willl the projected image on the screen (yes its the same distance away, everything else is the same) now be 6 feet by 6 feet? Is there a formula to figure this out using variables like, distance from the wall, focal length, film size etc.?

 

[ferider]

Yes.

Distance_to_wall / width_of_projected_image = focal_length / slide_width.

Everything else being equal, half the focal length leads to twice the image width.

 

[semi-ambivalent]

Yes.

Distance_to_wall / width_of_projected_image = focal_length / slide_width.

Everything else being equal, half the focal length leads to twice the image width.

2x the height too, yes? Then 4x the area, and 1/4 the image brightness.

 

[68degrees]

Interesting. But why brightness? The distance to the wall hasnt changed?

 

[Rob-F]

2x the height too, yes? Then 4x the area, and 1/4 the image brightness.

Yes, that is right.

 

[68degrees]

So if double the distance away from the wall, does that double the size of the projected image?

 

[68degrees]

Distance_to_wall / width_of_projected_image = focal_length / slide_width..

what would the formula for width of projected image?

Lets say I wanted an 2 meters wide image.

distance to wall is 5 meters

slide width is 60mm

what focal length and what would the formula be to solve focal length?

 

[Rob-F]

Here's how I do it. I make it simple, no need to remember a formula. I divide the distance from projector to screen by the focal length of the lens. The quotient is then the number of focal lengths from projector to screen. It is also the number of times the image will be magnified, from slide to screen. Then, by multiplying this number by the width of the slide, the result is the width of the projected image.

Ye Example:

We have a four inch lens, and the projector is ten feet from the screen. Ten feet is 120 inches (must always work in the same units). So 120 inches divided by 4 inches, gives us the distance from projector to screen, expressed in focal lengths: it's 30 focal lengths.

And that is also the magnification: the picture will be 30 times larger than the width of the slide. On a casual basis, for a rough approximation, we can say the 35mm slide is 1.5 inches wide. Really, it's a smidge less, but this is just for a casual quick estimate. So: 30 X 1.5" equals 45 inches of picture width.

If there is any interest, I can show how to calculate the picture width very accurately. But this method is intuitive and easy to remember, and is "close enough for government work," as we used to say in the U.S.A.F.

 

[68degrees]

Here's how I do it. I make it simple, no need to remember a formula. I divide the distance from projector to screen by the focal length of the lens. The quotient is then the number of focal lengths from projector to screen. It is also the number of times the image will be magnified, from slide to screen. Then, by multiplying this number by the width of the slide, the result is the width of the projected image.

Ye Example:

We have a four inch lens, and the projector is ten feet from the screen. Ten feet is 120 inches (must always work in the same units). So 120 inches divided by 4 inches, gives us the distance from projector to screen, expressed in focal lengths: it's 30 focal lengths.

And that is also the magnification: the picture will be 30 times larger than the width of the slide. On a casual basis, for a rough approximation, we can say the 35mm slide is 1.5 inches wide. Really, it's a smidge less, but this is just for a casual quick estimate. So: 30 X 1.5" equals 45 inches of picture width.

If there is any interest, I can show how to calculate the picture width very accurately. But this method is intuitive and easy to remember, and is "close enough for government work," as we used to say in the U.S.A.F.

Thats great thanks a lot. I just spent 2 hours making a spreadsheet but what you have provided is much simpler. So easy.

 

[semi-ambivalent]

Interesting. But why brightness? The distance to the wall hasnt changed?

The illumination thrown by the lamp hasn't changed either but, because of the shorter lens, that same 'quantity' of illumination is now spread out over four times the area: (2 times original height) by (two times original width).

s-a