Figuring Out Focal Length and Object Distance

[jesuslovesu]

Homework Statement

An object is located to the left of a thin lens. The image is upright, and the distance between the object and image is 20 cm. What are the focal length and object distance if the lens is 2 times the hieght of the object?

Homework Equations

1/p + 1/q = 1/f

The Attempt at a Solution

I actually know how to get the focal length and object distance, I'm just having a bit of difficulty with an initial step.

If the image and object are 20 cm apart. How do I determine what the relationship between them is? I know the image location (q) is negative and the object (p) is positive. According to my book, I should use 20 cm = -q - p. That isn't immediately obvious to me, I would have thought doing something like 20 = p - q or 20 = p + q... can anyone explain why it's -q - p for the distance differential?

 

[anathema]

Hi there!

I would like to clarify that the equation you are using, 1/p + 1/q = 1/f, is known as the thin lens equation. This equation relates the object distance (p), image distance (q), and focal length (f) of a thin lens.

To answer your question, the sign convention for distances in optics is as follows: distances to the left of the lens are negative, and distances to the right of the lens are positive. This means that the object distance (p) is negative since it is to the left of the lens, while the image distance (q) is positive as it is to the right of the lens.

In this case, we know that the object and image are 20 cm apart, so the image distance (q) is -20 cm (negative because it is to the left of the lens). To find the object distance (p), we can rearrange the thin lens equation to solve for p: p = qf/(q-f). Substituting in the values we know, we get p = (-20 cm)(f)/(f-(-20 cm)). Simplifying, we get p = -20f/(f+20).

We also know that the height of the lens is 2 times the height of the object, so we can set up a ratio using the object and lens heights: (object height)/(lens height) = (object distance)/(focal length). Substituting in the values we know, we get (object height)/(2*object height) = (-20f/(f+20))/f. Simplifying, we get 1/2 = -20/(f+20). Cross-multiplying, we get f = -40 cm.

Therefore, the focal length is -40 cm and the object distance is -40 cm. This means that the lens must be a diverging lens, as the focal length is negative, and the object is located at a distance of 40 cm to the left of the lens.

I hope this helps clarify your initial confusion. Let me know if you have any further questions. Keep up the good work in your studies!

 

[m2003]

I understand your confusion with the given equation. The distance between the object and the image is usually represented as the absolute value of the difference between their positions, which can be written as |p-q|. However, in this case, the negative sign in front of q indicates that the image is located on the opposite side of the lens from the object. Therefore, the distance between them is actually the sum of their positions, which can be written as -(p+q). This is why the equation given is -q-p.

To solve for the focal length and object distance, we can use the thin lens equation, 1/p + 1/q = 1/f, where p is the object distance, q is the image distance, and f is the focal length. We can rearrange this equation to solve for p and q:

1/p = 1/f - 1/q
p = 1/(1/f - 1/q)
p = qf/(q-f)

Substituting in the given information, where q = -20 cm and f = 2p, we get:

p = (-20 cm)(2p)/(-20 cm - 2p)
p = (-40p cm)/(2p - 20)
p = (-40p cm)/(2(p-10))

This gives us the object distance in terms of the focal length and the height of the object. We can also solve for the focal length by setting p = 2h, where h is the height of the object:

2h = (-40p cm)/(2(p-10))
2h = (-40(2h) cm)/(2(2h-10))
2h = (-80h cm)/(4h-20)
2h(4h-20) = -80h cm
8h^2 - 40h = -80h cm
8h^2 + 40h = 0
h(8h + 40) = 0

Since h cannot equal 0, the only solution for this equation is when 8h + 40 = 0, which gives us h = -5 cm. Therefore, the height of the object is -5 cm and the focal length is 2 times that, giving us a focal length of -10 cm.

To summarize, the focal length is -10 cm and the object is located at a distance of -5 cm from the