A microscope is sharply focused on a mark at 5.7 cm. If a parallel plate of glass is placed over the mark, Where will the focus be if the index of refraction of the glass is 1.52?

University Physics Volume 3
17th Edition
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:William Moebs, Jeff Sanny
Chapter2: Geometric Optics And Image Formation
Section: Chapter Questions
Problem 49P: Derive the spherical interface equation for refraction at a concave surface. (Hint: Follow the...
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A case problem is attached.

1. A microscope is sharply focused on a mark at 5.7 cm. If a parallel plate of glass is placed over the mark, Where will the focus be if the index of refraction of the glass is 1.52?

APPARENT DEPTH
Formula:
Apparent Depth = n2
Actual Depth
n1
Where:
Apparent Depth: is the distance due to refraction
Actual Depth: originates distance from the surface
n2: index of refraction of the 2d medium
n1: index of refraction of the 1ª medium (where the incident ray comes from)
Example Case Problem
Given:
Actual Depth: 6 inches
n2= 1.33 water
n1- 1.0 air
ni1.0 air
na-1.33 water
Transcribed Image Text:APPARENT DEPTH Formula: Apparent Depth = n2 Actual Depth n1 Where: Apparent Depth: is the distance due to refraction Actual Depth: originates distance from the surface n2: index of refraction of the 2d medium n1: index of refraction of the 1ª medium (where the incident ray comes from) Example Case Problem Given: Actual Depth: 6 inches n2= 1.33 water n1- 1.0 air ni1.0 air na-1.33 water
Example Case Problem
Given:
Actual Depth: 6 inches
n2= 1.33 water
n1- 1.0 air
ns-1.0 air
na1.33 water
apparent depth
na
ni-1.0 air
actual depth
ni
6
na-1.33 water
1.33
6"
1.0
1.0x
7.98
Therefore:
1.0
1.0
ni-1.0 air
x = 7.98"
6
n2-1.33 water
n=1.0 air
7.98
n1.33 water
7.98
ni-1.33 water
Rarer to denser object is farther
than where it is
4.51"
na1.0 air Denser to rarer object is percelved
nearer to the surface than where it is
located
Transcribed Image Text:Example Case Problem Given: Actual Depth: 6 inches n2= 1.33 water n1- 1.0 air ns-1.0 air na1.33 water apparent depth na ni-1.0 air actual depth ni 6 na-1.33 water 1.33 6" 1.0 1.0x 7.98 Therefore: 1.0 1.0 ni-1.0 air x = 7.98" 6 n2-1.33 water n=1.0 air 7.98 n1.33 water 7.98 ni-1.33 water Rarer to denser object is farther than where it is 4.51" na1.0 air Denser to rarer object is percelved nearer to the surface than where it is located
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