How Does the Focal Length of a Lens Vary with Red and Green Light?

In summary, the conversation discusses using a graphical approach to determine the difference in focal length of a lens when used with red and green light. The equation 1/u + 1/v = 1/f is mentioned as well as the inverse relationship between refractive index and wavelength. The speaker also mentions taking measurements and using an LED, lens, white block, and metre ruler to project the image. They ask for help in understanding the results they obtained and if anyone has done something similar before.
  • #1
Ahem
9
0
"To use a graphical approach to determine whether there is a difference between the focal length of a lens when used with red and green light."

What would I expect to see?

What sort of results do you think I will get?

What would be the variable?

I need to use the equation 1/u + 1/v = 1/f can anyone help in explaining it?

Has anyone done anything similar before?

Thanks for any help.
 
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  • #2
I can't remember the exact equation, but I do recall that the refractive index, n, of an object is inversely proportional to the wavelength (i.e. as the wavelength decreases, the amount the light is refracted by increases).

Have you been given any more information then that that you've given in your post? Because, otherwise, I can't see how to do more then give a vague approximation of what would happen.
 
  • #3
Well, using a suitable lens I had to take measurements over a range of object and image distances to help me plot the graph. I've got some results, but I don't know if they're correct or suitable.


I'm not sure what you mean in the first half of your post.
 
  • #4
My results are all ****ty, say for instance I wanted to make up some results. I'm using an LED with a green and red light, a lens, a white block (improvised as a screen to project the image -red or green light-) and a metre ruler. What sort of results do you think I would get or see?
 
  • #5
Anybody??
 

FAQ: How Does the Focal Length of a Lens Vary with Red and Green Light?

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