BHS 120.2 – Optics II
Notetaker: Fatima Ibrahim
Date: 02/6/2013, 2 nd hour
Page1
R EVIEW : T HICK L ENSES
Important Formulas:
Back vertex power: P v
’
Back vertex focal length: f v
’ = n image
/P v
’
Front vertex power: P v
Front vertex focal length: f v
= -n object
/P v
Equivalent power: P e
Primary Equivalent focal length: f e
= -n object
/P e
Secondary Equivalent focal length: f e
’ = n image
/P e
’
Primary Principle Plane: e = f v
- f e
Second Principle Plane: e’ = f v
’ - f e
’
Convergent lens: B > E > F > N
Divergent lens: N >B > E > F
*Light is refracted at the first principle plane and goes through the second principle plane to form an image.
Newton’s Relation : xx’ = f 2
Clicker Question 1: For the meniscus lens shown below, to the respective lens surface?
1. APF
2. PPF
will be located closer
The APF will be closer to the lens surface because of the object medium is less than the image medium. Since the media are different, B > E > F > N can’t be used to compare the APF and
PPF. Back and front vertex power won’t differ much, so the refractive index will cause the PPF to become longer. If the back and front vertex power were very different, then this generalization would not work, but for the most part this holds true. The larger refractive index would result in a longer focal length. So APF is closer.

BHS 120.2 – Optics II
Notetaker: Fatima Ibrahim
Date: 2/6/2013, 2 nd hour
Page2
Clicker Question 2:
The principle planes for the plus meniscus lens shown below will be shifted to the
1. Left
2. Right
side of the lens.
Principle planes shift to the steeper (stronger) surface. In this case, the anterior surface is steeper and stronger. So the principle planes will shift to the left.
Clicker Question 3:
The n odal points will shift
1. Anteriorly
2. Posteriorly
from their respective planes.
Nodal points usually align with principle planes. Since the medium for the object and image differs, the nodal points will shift. Nodal points follow the denser medium. Here the denser medium is 1.33 for the image. So the nodal points shift posteriorly.
Clicker Question 4: For the meniscus lens shown below, select the correct statement.
1. Front vertex focal length < 1o Equivalent focal length
2. Front vertex focal length > 1o Equivalent focal length

BHS 120.2 – Optics II
Notetaker: Fatima Ibrahim
Date: 02/6/2013, 2 nd hour
Page3
This question is comparing f v
to f e
, two lengths that share the same refractive index.
Using this B > E > F > N, the f e
must be shorter than the f v
. In other words, the front vertex focal length > 1o Equivalent focal length.
Clicker Question 5: The principle planes for the plus meniscus lens shown below will be shifted
of the lens.
1. Right
2. Left
Principle planes shift to the steeper (stronger) surface. In this case, the posterior surface is steeper and stronger. So the principle planes will shift to the right.
Clicker Question 6: T he nodal points will move from their respective principle planes.
1. Posteriorly
2. Anteriorly

BHS 120.2 – Optics II
Notetaker: Fatima Ibrahim
Date: 2/6/2013, 2 nd hour
Page4
Nodal points follow the denser medium. Here the denser medium is 1.33 for the object. This lens is a minus meniscus lens and the APF is on the right side, so the nodal points shift posteriorly.
Clicker Question 7:
For the
given lens, the principle planes will be shifted
1. Anteriorly (to the left side)
2. Posteriorly (to the right side)
3. Not enough information available at this point.
Principle planes shift to the steeper (stronger) surface. From the given front and back vertex power, the posterior surface is stronger. So the principle planes will shift to the right.
Clicker Question 8:
The nodal points are located closer to the posterior lens surface
1. True
2. False

BHS 120.2 – Optics II
Notetaker: Fatima Ibrahim
Date: 02/6/2013, 2 nd hour
Page5
Nodal points are located closer to the posterior surface since the posterior surface of the lens is stronger. Since the back vertex is stronger, it’s a minus lens and P
2
> P
1
. Since no medium is mentioned, we assume the medium is air. So the nodal points are coincident with the principle planes. So like the principle planes, the nodal points are closer to the posterior lens surface.
Sample Problem:
Given :
2.00cm thick polycarbonate meniscus lens
P v
= -3.58D
P e
= -3.85 D
P v
’ = -3.95D
Clicker Question 9: The 1o principle plane (H) is located __________ 1st lens surface
Use formula: e = f v
- f e e = (-1/-3.58) - (-1/-3.85) e = +1.925 cm
Clicker Question 10:
The 2o principle plane (H’) is located __________ 2nd lens surface
Use formula
: e’ = f v
’ - f e
’ e’ = (1/-3.95) - (1/-3.85) e’ = + 0.65 cm
Newton’s Relation
Newton’s relation uses the formula xx’ = f 2 , to locate objects and images with respect to focal points. The relation assumes that the primary and secondary focal lengths are equal for a thin lens.

BHS 120.2 – Optics II
Notetaker: Fatima Ibrahim
Extra-focal distances: x and x’ x = distance from object to APF x’ = distance from image to PPF
Date: 2/6/2013, 2 nd hour
Page6
Clicker Questio n 11:
An object is located 10.00 cm anterior to the real APF of a Cr-39 thin lens.
Its image is formed 40.00 cm posterior to the real PPF. Find the nominal power of the lens?
xx’= f 2
(0.10 m) (0.40 m) = f f = | 0.2 m |
2 = 0.04
P
P
N
N
= n surround
/|f|
= 1.00 / 0.20 m  |5.00 D|
P
N
=+5.00D since lens has real APF and a real PPF.
Clicker Question 12: An object is located 15.00 cm from the virtual APF of a polycarbonate thin lens. An image forms 20.00 cm from the virtual PPF. Find the nominal power.
xx’= f 2
(0.15 m) (0.20 m) = f 2 = 0.03 f = | 0.17 m |
P
N
= n surround
/|f|
P
N
= 1.00 / 0.17 m  |5.77 D|
P
N
= - 5.77D since lens has virtual APF and a virtual PPF.