Geometric Optics

Question 1

Describe what lenses are.

Answer 1

A tool used to bring light to a fixed focal point. There are 2 types of lenses: Convex and concave lenses

Question 2

Concave lenses 
A. Converge light rays 
B. Diffraction light rays 
C. Reflect light rays
D. Refract light rays

Answer 2

Concave lenses: these lenses cave in, focal points are on equal length on either sides of the lenses. These lenses tend to create an image at the front of the lens and therefore virtual images

Question 3

Describe Convex Lenses

Answer 3

These are lenses that converge light rays at a focal point on the other side of the lens. These images produced are real, inverted images. Much like concave lenses, these have 2 focal lengths as well.

Question 4

What is the thin lens assumption?

Answer 4

The thin lens assumption is an assumption that assumes the lens is thin enough that the angles of incidence and refraction are so small, they do not affect the travel of the light rays and therefore are negligible. Lens is considered to be thin if its thickness t is much less than the radii of curvature of both surfaces, in this manner the rays may be considered to bend once at the center of the lens only.

Question 5

Describe the center of curvature of a lens.

Answer 5

In lens, there are no center of curvatures like parabolic mirrors. 2f is still present, but this is not the same as the center of curvature as it is not the lense’s radius of curvature of each of the curves

Question 6

An object is placed further that the 2f position in front of a convex lens. Where does this image form? Describe the image.

Answer 6

Steps:
Pick the 2 rays again: (1) ray parallel to the principal axis (2) ray through the focal point on its respective side

Results: Real image, Inverted, smaller than original and closer to lens than the object
Note: If you are unsure of the distance the image will form, draw it out

Question 7

You place a plant directly in the 2f position in front of a converging lens. Where does this image form? Describe the image.

Answer 7

Pick the 2 rays again: (1) ray parallel to the principal axis (2) ray through the focal point on its respective side
Results: Image, real, inverted, image is about the same distance as the object is away from the lens (2f)
Note: If you are unsure of the distance the image will form, draw it out

Question 8

Mnemonic to remember converging and diverging lens

Answer 8

Converging = convex

Therefore the diverging lens is concave

Question 9

The rule of thumb for drawing out rays in front of ray diagrams are…

Answer 9

Draw 2 rays from the object to the optic device:
1) ray parallel to the principal axis, enters the optic device, bends, and then travels to directly perpendicular from the optic device to the focal point (if there is one)

(2) ray through the focal point on the same side of the object, enters the optic device and then travels straight parallel to the principal axis

Question 10

You’re working on a convex lens optics problem and you’re at the step where you’re figuring out light ray travel. You note the object is directly above the focal length. How does this disrupt your diagram. What modifications need to be done to correctly figure this problem out.

Answer 10

The ray that starts parallel to the principal axis remains the same, however the ray that travels through the focal point changes. This ray is not refracted, but instead goes straight through the lens without being refracted. In this manner, this object’s image is diffracted and no image is formed.

Question 11

On a converging lens optics problems, you note the object is anterior to the focal point in front of the lens. How does this change the rays of your ray diagram sketch?

Answer 11

Pick the 2 rays again: (1) ray parallel to the principal axis enters the optic device, bends, and then travels to directly perpendicular from the optic device to the focal point
(2) ray traveling as it has traveled through the focal point (represented with dotted lines behind the object) up to the mirror and travel parallel

Results: No real image is formed on the other side of the lens
These rays do appear to act as if they are diverging from some point (this point is on the other side lens where the object

Question 12

You’re setting up a diverging lens system and you start to sketch a ray diagram. What 2 lines are you supposed to draw out?

Answer 12

(1) Ray 1 from object run parallel to the principal axis, hits the lens and diffracts AWAY from the focal point
(2) Ray 2 from object runs perpendicular to the principal axis, hits the lens and continues in this pathway direction away from the focal point

Question 13

Vertical Angles of triangles are

Answer 13

Are opposite angles of 2 triangles. They articulate to one another have the same degree of angles

These angles are on the opposite sides of lines that intersecting

Question 14

Alternate Interior Angles are

Answer 14

2 angles of parallel lined triangles. These 2 angles are opposite and diagonal to one another and share the same degree in angles.

Question 15

True or False - The do variable in the thin lens equation describes the distance of the object from the focal point to the object

Answer 15

False, do describes the distance of the object from the center of the lens.
1/f = 1/do + 1/di

Question 16

Di of thin lens equation
A. Distance of the image to the object
B. Distance of the image to the lens 
C. Distance of the image to the center of curvature 
D. Distance of image to focal point

Answer 16

B. Distance of the image to the lens

1/f = 1/do + 1/di

Question 17

Through your glasses an image of a tree is formed 5 mm away from the glasses. If this image formed is 2cm tall and the tree is 15m away, how tall is the tree?

Answer 17

M = -di/do = hi/ho
5e-3 m / 15m = 2e-2m / ho
ho = 2e-2m(15m)/5e-3m
ho = 60 m

Question 18

In terms of lenses, what determines the sign of the focal point?

Answer 18

The signs for this value matters based on the lens type you have, not the direction you choose to measure
Convex lens AKA converging lens - f is ALWAYS positive
Concave Lens AKA Diverging lens - f is ALWAYS negative

Question 19

When is object distance negative?

Answer 19

do - object distance. Measured from the center of the lens to the object
Object distance is ALWAYS positive no matter the type of lens, as long as you are dealing with one lens
Exception: when there are multiple lens involved in the system - this can lead to a negative object distance in some cases

Question 20

When is di positive in terms of lenses?

Answer 20

When the image is produced on the opposite side from the object - Therefore object and image is not on the same side - di is ALWAYS positive (this is regardless of concave or convex)

Question 21

rue or False - The distance of the image is always positive when the lenses are concave.

Answer 21

False. When the image is produced on the same side of the object, di is ALWAYS negative (regardless of concave or convex lenses)

Question 22

In working on an optics lens problem with an inverted object, you come to find that the magnification is positive. What is the image? What type of lens does this usually imply?

Answer 22

M = - di/do
When M = + value, means that the object did not flip, therefore if the image is right side up, it will remain right side up image and vice versa.
Lens that tends to have a + M value are concave lens
Note: the actual value of M, and therefore the size of the image will be determined by the meters of the object and images.

Question 23

You see that your patient is a farsighted individual and wears convex glasses. Immediately, what can you say about the magnification of images with these glasses?

Answer 23

Convex lenses are associated with inverted, real images. This means that the M value will be positive

Note: the actual value of M, and therefore the size of the image will be determined by the meters of the object and images.

Question 24

How tall is the image if the object is 24cm from the focal length. Note: the focal length from the concave lens is 8cm.

Answer 24

Thin lens: 1/f = 1/do + 1/di
Magnification equation: M = -di/do [don’t know M, can’t use equation yet]

1/di = 1/f - 1/do
f = - because this is a concave lens => - 8 cm
do = + 24cm

1/di = -1/8 cm - 1/24cm
1/di = -4/24 cm
1/di = -1/6 cm 
-di = 6cm
di = -6cm

Question 25

Compare and contrast the different types of concave and convex lenses

Answer 25

Convex and concaves have both of the following:
Double sided lenses: both sides of the lenses are concave or convex based on the type of lense
Plano lens: One side of the lens is partially convex/concave
Concave/convex meniscus: One side of the lens is more concave/convex respectively

Question 26

As you scribe for Dr. Huygen, you hear the patient’s diopters are
L: -0.25 D
R: -0.75 D
What do these units mean?

Answer 26

P = 1/f
When power is in terms of m-1 this is known as diopters, D. Diopters determine the strength of the glasses needed to correct the patients eyesight in order for the image to form perfectly at the back of the retina

Question 27

When a patient wears prescription glasses, these glasses create a multi lens system. Assuming that the cornea and lens act as one lens, describe how you would solve for the image created by an object in front of your glasses.

Answer 27

Note: Work on one lens at a time
S1: Ray diagram of object through glasses. This forms an image behind the glasses.
S2: the image behind the glasses functions like the object for the second lens system (cornea and lens), this forms the second image behind this system and then converges back at the retina

Question 28

An image is formed 18 cm behind the first lens. Now imagine this image is refracted by the second concave lens, which has a focal length of 10cm. What is the image distance formed? [Note: the first lens and second lens are 33 cm apart]

Answer 28

f - is negative with concave lenses
do - is always positive, unless there are multiple lens systems. In this case it should be positive still though. di from lens 1 is 18cm from lens 1. HAVE TO subtract this value from lens #2 in order to find the distance of the image made to get the do = 33-18 = 15cm
di - should be negative after you solve because concave lenses creates images on the same side of the object. Do not attempt to plug in signs for di yet as it is the unknown [note: this lens will create an image of the created image from lens #1 on the same side!!!]

1/di = -1/f - 1/do
1/di = -1/10cm - 1/15cm 
di = -6cm

Question 29

You have a multiple lens system, where the first lens converges the image and creates an image half the size of the original object. The second lens system, which takes this image and diverges the image ⅖ smaller than the image of the first. What is the total magnification?

Answer 29

Total Magnification of the 2 lenses system:
Mtotal = M1M2… n
Therefore in this example: Mtotal = (-½)*(⅖) =
-2/10 = -⅕

Question 30

In the Lab, your first object is 10 inches high, the first image is 5 inches high, and the second image is 2 inches high. Your lab instructor asks you to calculate the overall magnification for the 2 lens set up. What is the answer?

A. 0.8
B. 0.5
C. 0.2
D. 0.1

Answer 30

di/do = hi/ho

Overall Magnification - Mtotal = M1M2… n
10 -> 5 ; this is a ½ magnification
5 -> 2 ; this is >½ magnification
Therefore Mtotal = (½)(½) = ¼ = 0.25
Answer C. 0.2 is the closest answer and remember that we rounded up

Question 31

What property of the lense can you change to cause it to have a more powerful strength? Why?

Answer 31

Remember that P = 1/f
the shorter the f, the more power the lens is. When the focal length is further away, it doesn’t bend light as much. Therefore it has less power in light bending. With f closer to the lens, it bends light more powerfully

Question 32

Spherical lenses create a spherical aberration. What is this?

Answer 32

This phenomenon occurs when parallel light rays on top at the edge of the lens are bent more than the others, meaning that it’ll before the focal point
Parallel light rays in the middle are bent less and converge at a point behind the focal point. These create a range/blur of light convergence known as spherical aberration.
This is an inherent problem with spherical lenses

Question 33

Why does spherical aberration occur?

Answer 33

The angle lines of light rays are small in thin lenses. When angles are small, sin θ = θ. With this assumption, we can assume that all rays go to the focal point
But with an increase in thickness, this changes the angle, and this leads to inherent problems

Question 34

Explain how color aberrations can exist in visible light

Answer 34

This aberration, or difference in convergence of light at different points, can occur differently within different frequency light.

Red experiences a smaller index of refraction. This means that these red rays will be bent less and will be more posterior to the focal point than it is supposed to

Blue rays - a higher frequency of light - experiences more bending because it has a higher index of refraction. This means that these lights will converge anterior to the focal point

Question 35

What is the problem the eye experiences because its lenses: cornea and lens, are convex lenses?

Answer 35

Both of these mean that these lenses converge light into a focal point at the back of the eye: the retina
However! Convex lenses create inverted images, this means that everything captured by the eye is upside down. (this is corrected by the brain though!)

Question 36

Humans in prehistoric times were fierce hunters. Along with their agile bodies, they were also able to quickly and sharply see objects. What structure in the eye allows the quick changes in vision in order for the lens to clearly focus images to the back of the brain?

Answer 36

The ciliary muscles
Note: When the ciliary muscles can’t compensate for this difference, usually the image will be formed posterior to the focal point, this leads to a blur image

Question 37

How does the ray model allow solving lens problems.

Answer 37

Ray model - light travels as light particles in a straight pathway. Assuming that light travels in a straight line allows us to depict essential rays that demonstrate if these rays from an object will converge or diverge

Question 38

In reflection, describe the relationship between the angle of incidence and the angle of refraction.

Answer 38

The angles of incidence equals to the angle of reflection - this theory is called the law of reflection

Question 39

In refraction, describe the relationship between the angle of incidence and the angle of refraction.

Answer 39

The angle of refraction - the angle of light ray after it passes from one medium to another - is related to the angle of incidence through snell’s law - n1sinθ1 = n2sinθ2. These angles are determined by the medium (this affects the n value)

When a ray enters a medium with an increased index refraction, the angle of refraction decreases, therefore the angle of refraction is less than the angle of incidence. And vice versa

Question 40

A light ray coming through the window exits the glass and enters your studio apartment. Describe the relationship of angle of incidence to the angle of incidence.

Answer 40

Because the index of refraction is higher in glass, the angle of incidence (the incidence ray (from glass to air)) is smaller than the angle of refraction.
n1sinθ1 = n2sinθ2

Question 41

If an object is on the same side of a lens as the image, this is considered to be a \_\_\_\_\_\_\_\_ image. If an object is on the opposite side of a lens as the image, this is considered to be a \_\_\_\_\_\_\_\_ image.
(A) virtual, virtual
(B) virtual, real
(C) real, real
(D) real, virtual

Answer 41

(B) virtual, real
If an object is on the same side of a lens as the image, this is considered to be a virtual image. If an object is on the opposite side of a lens as the image, this is considered to be a real image.

Question 42

When is an image considered Upright vs. Inverted?

Answer 42

An image is considered to be Upright when it is above the horizonal axis/plane (the principal axis). An image is considered to be Inverted when it is below the horizonal axis/plane (the principal axis).

Question 43

An object is at a distance of 30 cm from a convex lens with a focal length of 20 cm. What is the image distance (in cm) for this lens? Is the image inverted or upright? Is the image real or virtual?

(A) 90
(B) 60
(C) -60
(D) -90

Answer 43

(B) 60
1/f = 1/o + 1/i
1/20 = 1/30 + 1/i
i = 60
The image distance is 60 cm. The image is inverted and real. 

NOTE: You can use the thin lens equation, but I find that it is much easier to simply draw it out and estimate the answer since exact numbers are not a requirement on the MCAT.