.Drawing Ray Diagrams: Tips & Tricks for Solving Problems

[GingerBread27]

I'm having a lot of trouble drawing ray diagrams (this is for a lab assignment). If anyone could give me some tips or links to websites that are helpful in teaching how to draw ray diagrams that would be great. One example of a problem is: A 1-cm tall arrow (the object) is placed 2 cm to the left of a converging lens having a 7 cm focal length. Show all three pricipal rays. Is the image real/virtual?Erect of inverted? Measuring from diagram determine the location and magnification of the image.

Now I think that I draw an upright arrow, 1 cm tall, to the left of the lens and then 7 cm to the right of the lens i draw another arrow? Is it upside down, larger/smaller? I have no idea what I'm doing and the write up for the lab is not helpful. Please help! thanks

 

[Doc Al]

You draw the three principle rays emanating from the tip of the 1-cm arrow you drew (which is your object). If you draw them correctly, they will meet at the point where the image of that arrow tip is. You'll be able to tell whether the image is real, inverted, magnified, just by looking at your (carefully drawn) diagram.

Here's a site that tells you how to draw those rays: http://www.glenbrook.k12.il.us/gbssci/phys/Class/refrn/u14l5da.html

 

[GingerBread27]

For the problem I posted above would it be correct to do:

(1/do)+(1/di)=(1/f)=== (1/2)+(1/di)=(1/7), to get a di=-14/5=-2.8 cm, putting the image at 2.8 cm to the right of the lens?
Or am I completely clueless as to what focal length truly means, I thought it meant that the image was 7 cm from the lens.

 

[Berislav]

Or am I completely clueless as to what focal length truly means, I thought it meant that the image was 7 cm from the lens.

No, that's not what it means. The focal length is the property of the lens itself.

to get a di=-14/5=-2.8 cm, putting the image at 2.8 cm to the right of the lens?

Notice that $d_i$ is negative. This means that the image is virtual. So, the image is to the left of the lens.

 

[Doc Al]

For the problem I posted above would it be correct to do:

(1/do)+(1/di)=(1/f)=== (1/2)+(1/di)=(1/7), to get a di=-14/5=-2.8 cm, putting the image at 2.8 cm to the right of the lens?

Your use of the lens equation is correct, but your interpretation of the minus sign is not. (See Berislav's comments.) The sign convention says that a positive image distance is to the right of the lens; a negative image distance, to the left.

Or am I completely clueless as to what focal length truly means, I thought it meant that the image was 7 cm from the lens.

A focal length of 7cm means that parallel rays will be focused at the focal point, which is 7cm past the lens; it describes how much the lens converges (or diverges) the light that goes through it. (The shorter the focal length, the more powerful the lens.) Where the image is formed depends on (1) where you put the object and (2) the focal length of the lens. To find where the image is, use the lens equation (just as you did).