How Do You Calculate the Index of Refraction in This Microscopy Scenario?

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To calculate the index of refraction of the plastic in the microscopy scenario, the objective's adjustment of 0.40 cm when a 1.00-cm-thick piece of plastic is placed over the dot is crucial. The relationship between the angles of incidence and refraction can be expressed using the formula N = (sinA/sinP), which simplifies under the small-angle approximation to a ratio of heights. The horizontal distance from the center to where the ray exits the plastic is defined by the tangent of the angles involved. This analysis indicates that the index of refraction can be derived from the geometry and adjustments made during focusing. Understanding these relationships is essential for solving the problem accurately.
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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:

N=(sinA/sinP)=(5/3)*(cosA/cosP)..

I really need some help for this question...
 

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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.
 
oh, yes,...the approximation..
thank you very much.
 
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