|Feb2-06, 01:22 AM||#1|
need help on an optic question
the question is:
a mircroscope is focused on a black dot. when a 1.00-cm-thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.40 cm to bring the dot back into focus. what is the index of refraction of the plastic...
I spent a lot of time on this question, but still can not solve it.
These are what i got so far:
I really need some help for this question...
|Feb2-06, 06:13 PM||#2|
The horizontal distance along the top of the plastic,
from the center to the place the ray exits the plastic,
is r = h_green * tan(theta_1) = 1 cm * tan(theta_2).
This is supposed to be true for all rays (including rays closer to the center),
which can only be "true" in the small-angle-approximation
(sin(theta) = theta[radians] = tan(theta)).
In this approximation, your ratio of sines becomes a ratio of angles,
which becomes a ratio of tangents, which becomes a ratio of heights.
|Feb3-06, 02:47 AM||#3|
oh, yes,...the approximation..
thank you very much.
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