Understanding Thin Lens Equations: Solving for Image Distances and Focal Length

[badd99]

Homework Statement

I am given 5 distances to place the object relative to the lens. I then need to calculate each of the images distances for all 5 and record them. Once I have that done, I am to conduct a graph of 1/q vs 1/p. I then need to answer these questions:

1- Is the slope of your graph consistant with what you would expect from 1/q = -1/p + 1/f?

2-If your lens is placed at 15cm away (in front) of the object, where is the image? Is the image real or virtual? (To answer this question use the focal length you obtained from your graph).

Homework Equations

1/f = 1/q + 1/p

1/q = -1/p + 1/f

The Attempt at a Solution

Okay, so for number 1 - When I get the slope of my graph how do I compare it to that equation when I have 5 different image distances? Can I just pick one image distance and solve for -1/p since its in the form of x in y=mx+b?

2- Can I use 1/q = -1/p + 1/f and set p here to be 15cm? I would use my graph of 1/q vs 1/p to set the intercept to 1/f and solve for f to find the focal length and use in the equation with the 15cm. Also, should my focal lengths always be in meters with this equation? Also, if I solved with the values for this equation and 1/q came out to be negative would the image be virtual?

 

[Mindscrape]

Not sure whether q represents object or image since you didn't tell us. Q1 tells you to keep your focal length fixed, and then vary the object distance to see where the object images. So yeah, say whether it gives you an expected line or not.

You should use your graph to answer question 2, no calculation needed.