Lenses and Optics

Question 1

What is the strength of a lens, conceptually?

Answer 1

The strength of a lens (measured in diopters) is a measure of the len’s refractive power. The greater the strength, the more the light will bend.

Question 2

What is the formula for lens strength?

Answer 2

Strength = (1/Focal Length)

This yields the strength of a lens in diopters.

Question 3

How does the strength of a lens relate to focal length?

Answer 3

More powerful lenses have a shorter focal length. This is because the light needs to bend more at the shorter focal point.

Question 4

What do negative and positive values mean for lens strength?

Answer 4

Negative value: lens is diverging (i.e., concave)

Positive value: lens is converging (i.e., convex)

Question 5

How does a converging lens refract parallel rays of light?

Answer 5

A converging (convex) lens refracts parallel rays of light so they converge at a single point on the other side of the lens called the focal point.

Question 6

Is a converging lens convex or concave?

Answer 6

A converging lens is convex.

CONVerging = CONVex

Question 7

Is a diverging lens convex or concave?

Answer 7

Concave

Question 8

What is the sign of the focal length (F) for diverging and converging lenses?

Answer 8

Diverging (concave): negative

Converging (convex): positive

Question 9

What kind of image does a diverging lens produce?

Answer 9

Virtual, upright, and reduced. Note that virtual images are always upright!

So, this means that the image distance (Di) for diverging lenses is always negative.

Question 10

How does a diverging lens refract parallel rays of light?

Answer 10

A diverging (concave lens) spreads parallel rays of light away from its focal point.

Question 11

What is myopia and how do we treat it?

Answer 11

Myopia is nearsightedness and results when light from distance objects is focused in front of the retina. NEARsightedness = light focused in front of retina.

To treat myopia, we need to use a diverging (concave) lens. This will “push” the image back to the retina.

Question 12

What is hyperopia and how do we treat it?

Answer 12

Hyperopia is farsightedness and results when light focuses behind the retina. FARsightedness = light focused too FAR back.

To treat hyperopia, we need to use a converging lens that will focus the light sooner.

Question 13

Image distance (Di)

Answer 13

The image distance is the distance from the lens or mirror to the image.

Question 14

What does the sign of the image distance tell us?

Answer 14

Positive image distance: real image

Negative image distance: virtual image

Question 15

Object distance (Do)

Answer 15

The object distance is the distance from the object to the lens.

Question 16

How to solve for the focal length

Answer 16

Focal length = 0.5(R), where r is the radius of curvature.

Note that if we have a diverging (concave) lens, we need to make the focal length negative!

Question 17

What is chromatic aberration and how does light frequency relate?

Answer 17

Chromatic aberration refers to the formation of blurry images due to the effects of dispersion (light separation) through a lens.

Note that the refractive index of light increases with frequency, so high frequency light (e.g., violet) will refract more.

Question 18

What is spherical aberration?

Answer 18

Spherical aberration occurs in converging lenses and is due to the geometry to a spherical lens. Light rays near the edge of a spherical lens refract more than predicted due to its thickness, which can lead to multiple focal points. As a result, blurriness can occur.

Question 19

How to calculate magnification for lenses placed in series?

Answer 19

M total = M1 * M1 * … * Mn

Question 20

How can we fix spherical aberration?

Answer 20

Reduce the thickness of the lens.

Question 21

What kind of image does a convex (diverging) mirror produce?

Answer 21

Diverging mirrors always produce a virtual, upright image.

Question 22

What kind of lens is the eye?

Answer 22

The eye is a converging lens.

Question 23

What kind of image is produced in a converging lens or mirror if the object is placed within the focal length (i.e., O less than F)?

Answer 23

A upright and virtual image is formed.

Question 24

What kind of image is produced in a converging lens or mirror if the object is placed outside the focal length (i.e., O>F)?

Answer 24

An inverted and real image is formed.

Question 25

What kind of image is produced in a converging lens or mirror if the object distance equals the focal length (i.e. O=F)?

Answer 25

No image is produced.