Optics, slide projector

[studentteacher]

I think I'm just blanking on the algebra.

 

Problem: A slide projector (old book) needs to project a 98 cm high image of a 2cm tall slide. The screen is 320 cm from the slide.

 

What focal length does the lens need?

 

How far should the lens be from the slide?

 

So, image-height = 98 cm. object-height = 2 cm. object-distance + image-distance = 320 cm.

 

Formulas are 1/do + 1/di = 1/ f

 

and hi/ho = - di/do.

 

Should be able to find f, but can't.....

[LabRat1]

Do you know any formulas which my include magnification of the lens?

[studiot]

Do you know any formulas which my include magnification of the lens?

 

The OP actually did state one.

 

However, only two equations were stated and three are available from the question information.

 

Working it through I find that this does not work out to particularly simple numbers, but my figures place the object on the correct side of the focus.

 

Solution strategy

 

1) Form two equations in the two unknowns, d_i and d_o, and solve.

 

2) Substitute these values into the lens equation you have stated and solve for f.

 

Take care with the sign convention you are using with the negative sign in the magnification equation

What is its significance?

 

Post your working.

Edited March 14, 2017 by studiot

[studentteacher]

I can't get a sensible answer. I think the problem is that it is wrong to state "object-distance + image-distance = 320 cm." because one of the distances is going to be negative. Using substitution, I get that do = -6.67 and di then = 326.67, which is impossible.

 

I don't really understand presenting the problem this way. There must be a general relationship for distance between image and object, instead distance between object & image and lens.

Edited March 14, 2017 by studentteacher

[studiot]

I did say to watch out for the sign convention, yours is not the easiest.

 

I also said to post your working - it makes it a lot easier to help that way.

 

As it is all I can do is list the equations you should have, which you have not done.

 

So

 

Equation 1

 

d_o + d_i = 320

 

Equation 2

 

d_i/d_o = -1* (98/2) = 49

 

Note I have put the minus sign to a 1 here because neither the object nor the image distance is negative.

The -1 tells you that the image is inverted, but you should also know (the equations don't tell you) that the image is real and therefore the image distance is positive.

 

or d_i = 49d_o

 

or d_o = d_i/49

 

Substituting first for d_o and then for d_i yields

 

d_o = 6.4

 

d_i = 313.6

 

(These figures correctly add up to 320)

 

Note that the quantity (d_o + d_i) is called the 'throw' of the lens and is often given the symbol T.

 

Can you now use your third equation to calculate the focal length?

 

and show your working please.

Edited March 14, 2017 by studiot

[studentteacher]

I don't understand this part. How can it be correct to write -1 * (x) = x ?

 

d_i/d_o = -1* (98/2) = 49

 

Note I have put the minus sign to a 1 here because neither the object nor the image distance is negative.

The -1 tells you that the image is inverted, but you should also know (the equations don't tell you) that the image is real and therefore the image distance is positive.

or d_i = 49d_o

or d_o = d_i/49

 

My problem-solving was the same as yours, except I used d_i/d_o = - 49, because that's what the equation says to use.

Edited March 18, 2017 by studentteacher

[studiot]

I don't understand this part. How can it be correct to write -1 * (x) = x ?

 

My problem-solving was the same as yours, except I used d_i/d_o = - 49, because that's what the equation says to use.

 

Exactly and where did this equation come from?

 

I have asked you several times what sign convention you are using and tried to make it fit, but you have not replied.

 

Do you know what a sign convention is?

 

Do you understand that the image is real and what this means?

 

I can only help properly if you answer my questions - they are not tricks they are meant to help.

[studentteacher]

If object distance is positive it is in front of the lens. If image distance is positive, the image is behind the lens and real. (But, our text doesn't give a coherent definition of "image" or "front" or "behind." I assume "front" means where the light is coming from and "behind" means where the lens sends the light. How can an image be in front of a lens? Or an object behind it?)

 

In this case. the object (source of light) is the slide, and the image is "the screen". Since they are on opposite sides of the lens, they have the same sign, and the sign in the equation doesn't require modification. I guess.