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

  • Thread starter Thread starter zhen
  • Start date Start date
  • Tags Tags
    Optic
Click For Summary
SUMMARY

The discussion focuses on calculating the index of refraction of a plastic layer placed over a dot viewed through a microscope. The scenario involves a 1.00-cm-thick piece of plastic that requires a 0.40 cm adjustment in the microscope objective to regain focus. The formula presented for the index of refraction (N) is derived from the relationship between angles of incidence and refraction, specifically using the small-angle approximation. The key takeaway is that the index of refraction can be calculated using the relationship N = (sinA/sinP) = (5/3)*(cosA/cosP).

PREREQUISITES
  • Understanding of geometric optics principles
  • Familiarity with the small-angle approximation in trigonometry
  • Knowledge of the sine and cosine functions
  • Basic concepts of microscopy and optical adjustments
NEXT STEPS
  • Research the derivation of Snell's Law in optics
  • Study the small-angle approximation and its applications
  • Explore the principles of microscope optics and adjustments
  • Learn about the calculation of indices of refraction for different materials
USEFUL FOR

Students and professionals in optics, microscopy technicians, and anyone interested in understanding the principles of light refraction and its practical applications in microscopy.

zhen
Messages
48
Reaction score
0
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...
 

Attachments

  • optic.JPG
    optic.JPG
    8.5 KB · Views: 634
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
4K
Replies
1
Views
2K
Replies
3
Views
4K
  • · Replies 6 ·
Replies
6
Views
9K
Replies
1
Views
1K
Replies
2
Views
4K
Replies
4
Views
8K
Replies
1
Views
9K
  • · Replies 4 ·
Replies
4
Views
5K