Light & Optics

Question 1

What law gives the angle of reflection of a light ray from a flat surface?

Answer 1

θ_i = θ_r

This expression is the law of reflection, which states that the angle of incidence will equal the angle of reflection for a light ray striking a surface. Note that all angles are measured from the normal.

Question 2

A light ray strikes a pane of glass, making an angle of 35º with the surface. Find the angle of reflection.

Answer 2

θ_r = 55º

The angle of incidence always equals the angle of reflection. However, remember that the angles of incidence and reflection must be measured relative to the normal. If the angle made with the surface is 35º, the angle made with the normal will be (90º - 35º) = 55º. θ_i and θ_r are shown below.

Question 3

Define:

refraction

Answer 3

This occurs when a light ray crosses a boundary from one transparent medium to another.

When a light ray travels between media, it will bends either toward or away from the normal, depending on the properties of the media involved.

Question 4

Define:

index of refraction

Answer 4

The index of refraction (n) of a transparent medium is a unitless ratio of the speed of light in that medium compared to the speed of light in a vacuum.

Index of refraction is defined as n = c/v. Here, n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of light in the medium.

Question 5

What are the approximate indices of refraction of air, water, and glass, respectively?

Answer 5

• n_air = approximately 1
• n_water = approximately 1.33
• n_glass varies greatly, but 1.5 is a commonly-used value on the MCAT

Question 6

What features characterize substances with indices of refraction that are less than 1?

Answer 6

No substance has an n value that is less than 1.

Index of refraction is defined as c/v, where c is light’s speed in a vacuum and v is its speed in the relevant medium. Since light travels fastest in a vacuum, n is always equal to or greater than 1.

Question 7

How does the path of a light ray change when it crosses a boundary from air into glass?

Answer 7

It bends towards the normal of the surface.

Air has a lower index of refraction than glass. Whenever light moves from lower to higher index of refraction, it will angle toward the normal.

Question 8

How does the path of a light ray change when it crosses a boundary from water into air?

Answer 8

It bends away from the normal of the surface.

Water has a higher index of refraction than air. Whenever light moves from higher to lower index of refraction, it will angle away from the normal.

Question 9

What quantities are related by Snell’s law?

Answer 9

It gives the relationship between the angle of incidence and that of refraction for a light ray crossing the boundary between media. The equation is:

n₁sin(θ₁) = n₂sin(θ₂)

where
n₁ = index of refraction on the incident side
θ₁ = angle of incidence
n₂ = index of refraction on the refraction side
θ₂ = angle of refraction

Question 10

What is the critical angle (θ_c)?

Answer 10

It is the angle of incidence in a certain medium at which the angle of refraction into a new medium would reach 90º. Angles larger than the critical angle will result in total internal reflection.

A critical angle only exists when light is moving from a high to a low index of refraction. A classic example is refraction from diamond into air.

Question 11

What formula is used to calculate the critical angle (θ_c)? Assume that light is traveling from a medium with refractive index n₁ to a medium with index n₂.

Answer 11

θ_c = sin^-1(n₂/n₁)

This formula can be derived from Snell’s law:

n₁sin(θ_c) = n₂sin(90º)
n₁sin(θ_c) = n₂(1)
sin(θ_c) = n₂/n₁
θ_c = sin^-1(n₂/n₁)

Question 12

The utility of a fiber optic cable depends on its ability to trap light inside its own material for long distances. What property is likely involved in this example?

Answer 12

total internal reflection

In such cases, light rays stay entirely in the incident medium; no light refracts across the boundary. This only occurs if the n value of the cable is sufficiently higher than that of the outside material, and if the angle of incidence is large enough.

Question 13

Under what conditions does total internal reflection occur?

Answer 13

• Light must be traveling from high to low index of refraction.
• The angle of incidence of the light ray must exceed the critical angle of the interface.

Question 14

Define:

dispersion

Answer 14

This involves the refraction of different wavelengths of light at different angles.

The most common example of dispersion tested on the MCAT involves white light traveling through a prism. When dispersion occurs, the light splits into a spectrum of colors and spreads out at a range of angles.

Question 15

What process causes a beam of sunlight to separate into a rainbow of colors as it passes through a prism?

Answer 15

dispersion

Like all white light, sunlight is made up of all the colors of visible light mixed together. Different colors of light refract differently as they pass through the prism, so they are separated upon exiting the back surface.

Question 16

When white light is separated by a prism into its component colors, in what order do they appear?

Answer 16

• red
• orange
• yellow
• green
• blue
• indigo
• violet

This is easily remembered by the mnemonic ROYGBIV (Roy Gee Biv). Note that violet light, which travels slowly and has a high frequency, bends the most while red light bends the least.

Question 17

Define:

focal point

Answer 17

An optic’s focal point is the location where light rays parallel to its axis will cross after reflecting from, or refracting through, the optic.

An optic with a short focal point is strong and bends light drastically, while an optic with a long focal point is weak and affects light rays less significantly.

Question 18

In an optical system, what is an image?

Answer 18

It is an optical reproduction of a physical object, formed when light rays transmitted by the object are converged at a specific point by an optic or series of optics.

Many optics problems will involve calculating the location of a system’s image, given a particular object and a lens or mirror.

Question 19

What types of lens or mirror has a radius of curvature?

Answer 19

All spherical lenses and mirrors have radii of curvature.

Specifically, the radius of curvature of a mirror is the radius of the sphere from which the mirror or lens is a small portion.

Question 20

How does the radius of curvature of a lens or mirror compare to its focal length?

Answer 20

twice the focal length

In other words, focal length can be found by taking one-half of the radius of curvature.

Question 21

Lenses A and B are both converging lenses. How will the strength of the two lenses compare if the radius of curvature of Lens A is significantly larger than that of Lens B?

Answer 21

Lens B will be stronger than Lens A.

Since Lens A has a larger radius of curvature, it must also have a larger focal distance. The larger the focal length, the farther away from the lens the image will be formed; a weak lens will therefore have a large focal length. This also relates to the equation P = 1/f, which shows that power decreases as focal length increases.

Question 22

When is the image produced by an optical system real?

Answer 22

An optical system projects a real image when light rays reflected off or transmitted through the optic cross after the reflection or transmission.

Real images include the images on movie screens and the images made by the eye on the retina.

Question 23

When is the image produced by an optical system virtual?

Answer 23

An optical system projects a virtual image when light rays reflected off or transmitted through the optic diverging away from one another.

Virtual images include the images created by flat mirrors and magnifying glasses.

Question 24

In what cases will a virtual image be upright, or erect?

Answer 24

Any virtual image produced by any system will be upright.

Remember “IR” and “UV”; inverted images are always real, while upright images are always virtual.

Question 25

In what cases will a real image be upright, or erect?

Answer 25

No real image produced by a system will ever be upright. Real images are always inverted.

Remember “IR” and “UV”; inverted images are always real, while upright images are always virtual.

Question 26

What qualities define a converging mirror?

Answer 26

If any inbound light ray parallel to the mirror’s principal axis crosses that axis after reflecting off the mirror.

Converging mirrors are concave in shape.

Question 27

What is the focal length of a converging mirror?

Answer 27

It is the distance from the mirror to the focal point. Rays parallel to the optical axis cross at the focal point after reflecting off the mirror.

The focal length of any spherical mirror or lens is one-half of its radius of curvature.

Question 28

An optical system consists of an object and a single converging mirror. If the object is located between one and two focal lengths from the mirror, what are the properties of the final image?

Answer 28

The image is real, inverted, and magnified. It is also more distant from the mirror than the object is.

Note that converging lenses also produce real, inverted images when the object distance fits these specifications.

Question 29

An optical system consists of an object and a single converging mirror. If the object is located exactly one focal length from the mirror, what are the properties of the final image?

Answer 29

No image is formed.

Note that converging lenses also fail to produce an image when the object is positioned one focal length from the lens.

Question 30

An optical system consists of an object and a single converging mirror. If the object is located less than one focal length from the mirror, what are the properties of the final image?

Answer 30

The image is upright, virtual, and magnified. It is also more distant from the mirror than the object is.

Note that converging lenses also produce upright, virtual images when the object distance fits these specifications.

Question 31

What is the thin lens equation used to calculate?

Answer 31

It relates an optical system’s focal length to its object and image distance. This equation works for both lenses and mirrors.

The equation reads:

1/f = 1/o + 1/i

where
f = focal length
o = distance between optic and object
i = distance between optic and image

Question 32

What is the magnification of an optical system?

Answer 32

It is the ratio of the size of the image to the size of the object.

Magnification can be calculated from the image and object distance of the optical system. The magnification m is calculated as:

m = -(image distance/object distance)

Question 33

What does the sign of its magnification indicate about an image?

Answer 33

The sign of the magnification (positive or negative) indicates the direction that the image points.

If m > 0, the image is upright. If m < 0, the image is inverted.

Question 34

What does the magnitude of its magnification indicate about an image?

Answer 34

The magnitude of the magnification indicates the size of the image.

If |m| > 1, the image is magnified, or larger than the object. If |m| < 1, the image is reduced, or smaller than the object.

Question 35

For an optical system, the object and image distance are 30 cm and 60 cm, respectively. What is the magnification of the image?

Answer 35

m = -2

Remember, m = -i/o; so for this system:

m = -i/o = -(60/30) = -2

The image is inverted, and is twice as tall as the object.

Question 36

For an optical system, the object and image distance are 20 cm and -10 cm, respectively, What is the magnification of the image?

Answer 36

m = +½

Remember, m = -i/o; so for this system:

m = -i/o = -((-10)/20) = +½

The image is upright, and is half as tall as the object.

Question 37

What qualities define a diverging mirror?

Answer 37

A mirror is diverging if any inbound light ray which is parallel to the principal axis points away from the axis after reflecting off the mirror.

Diverging mirrors are convex in shape.

Question 38

What does the sign of an optic’s focal length indicate about its properties?

Answer 38

An optic’s focal length indicates the converging or diverging nature of the optic.

If f > 0, the optic is converging, while if f < 0, the optic is diverging. If f = infinity, the optic is planar.

Question 39

An optical system is made up of a single mirror, with an object distance of 30 cm and an image distance of -30 cm. What type is the mirror?

Answer 39

The mirror is planar. This is evident because f = infinity.

Given i and o, f can be calculated as:

1/f = 1/o + 1/i = 1/30 + 1/(-30) = 0

If 1/f = 0, f = infinity.

Planar mirrors produce virtual images, and i = -o.

Question 40

An optical system is made up of a single mirror, with an object distance of 15 cm and an image distance of 30 cm. What type is the mirror?

Answer 40

The mirror is converging, and f = 10 cm.

Given i and o, f can be calculated as:

1/f = 1/o + 1/i = 1/15 + 1/30 = 3/30
f = 30/3 = 10 cm

Only converging mirrors and lenses have positive focal lengths.

Question 41

What is the focal length of a diverging mirror?

Answer 41

The focal length is the distance from the mirror to the focal point. Virtual extensions of rays parallel to the principal axis cross at the focal point after reflecting off the mirror.

The focal length of any spherical mirror or lens is one-half of its radius of curvature.

Question 42

An optical system consists of an object and a single diverging mirror. What are the properties of the final image?

Answer 42

The image is virtual, reduced, and upright. It is also closer to the mirror than the object is.

Note that diverging lenses also always produce upright, virtual images.

Question 43

An optical system is made up of a single mirror, with an object distance of 30 cm and an image distance of -15 cm. What type is the mirror?

Answer 43

The mirror is diverging, and f = -30 cm.

Given i and o, f can be calculated as:

1/f = 1/o + 1/i = 1/30 + 1/(-15) = -1/30
f = -30 cm

A negative focal length generally correlates to a diverging system.

Question 44

What qualities define a converging lens?

Answer 44

A lens is converging if any inbound light ray which is parallel to the optical axis crosses the axis after refracting through the lens.

Converging lenses are convex in shape.

Question 45

What is the focal length of a converging lens?

Answer 45

The focal length is the distance from the lens to the focal point. Rays parallel to the principal axis cross at the focal point after refracting through the lens.

The focal length of any spherical mirror or lens is one-half of its radius of curvature.

Question 46

An optical system consists of an object and a single converging lens. If the object is located between one and two focal lengths from the lens, what are the properties of the final image?

Answer 46

The image is real, inverted, and magnified. It is also more distant from the lens than the object is.

Note that converging mirrors also produce real, inverted images when the object distance fits these specifications.

Question 47

An optical system consists of an object and a single converging lens. If the object is located exactly one focal length from the lens, what are the properties of the final image?

Answer 47

No image is formed.

Note that converging mirrors also fail to produce an image when the object is positioned one focal length from the mirror.

Question 48

An optical system consists of an object and a single converging lens. If the object is located less than one focal length from the lens, what are the properties of the final image?

Answer 48

The image is upright, virtual, and magnified. It is also more distant from the lens than the object is.

Note that converging mirrors also produce upright, virtual images when the object distance fits these specifications.

Question 49

An optical system is made up of a single lens, with an object distance of 20 cm and an image distance of 20 cm. What type is the lens?

Answer 49

The lens is converging, and f = 10 cm.

Given i and o, f can be calculated as:

1/f = 1/o + 1/i = 1/20 + 1/20 = 1/10
f = 10 cm

Only converging mirrors and lenses have positive focal lengths.

Question 50

What qualities define a diverging lens?

Answer 50

A lens is diverging if any inbound light ray which is parallel to the principal axis points away from the axis after refracting through the lens.

Diverging lenses are concave in shape.

Question 51

What is the focal length of a diverging lens?

Answer 51

The focal length is the distance from the lens to the focal point. Virtual extensions of rays parallel to the principal axis cross at the focal point after refracting through the lens.

The focal length of any spherical mirror or lens is one-half of its radius of curvature.

Question 52

An optical system consists of an object and a single diverging lens. What are the properties of the final image?

Answer 52

The image is virtual, reduced, and upright. It is also closer to the lens than the object is.

Note that diverging lenses also always produce upright, virtual images.

Question 53

An optical system is made up of a single lens, with an object distance of 45 cm and an image distance of -15 cm. What type is the lens?

Answer 53

The lens is diverging, and f = -22.5 cm.

Given i and o, f can be calculated as:

1/f = 1/o + 1/i = 1/45 + 1/(-15) = -2/45 = -1/22,5
f = -22.5 cm

A negative focal length generally correlates to a diverging system.

Question 54

What quantity is measured in diopters?

Answer 54

It measure the power, or strength, of optical elements.

Power is calculated by taking the reciprocal of the focal length:

P = 1/f

Focal lengths are typically reported in m, so 1 diopter is 1 m^-1. Note that you may also see optical measurements in cm.

Question 55

Which lens bends light more dramatically, a 2-diopter lens or an 8-diopter one?

Answer 55

The 8-diopter lens bends light more.

The shorter a lens’ focal length, the closer light converges on the optical axis after passing through the lens. The focal length of the 2-diopter lens is 1/2 m, while that of the 8-diopter lens is 1/8 m. The 8-diopter lens, with its shorter focal length, bends light more.

Question 56

If a 2-diopter lens is combined with a 3-diopter lens, what is the approximate strength of the combined lens system?

Answer 56

The combined lens system has a strength of approximately 5 diopters.

The diopter system is roughly additive; when multiple elements are combined, their overall diopter rating is roughly the sum of the strength of each individual element.

Question 57

Define:

chromatic aberration

Answer 57

Chromatic aberration is the characteristic colored halo around an image created by a lens.

This type of aberration occurs due to dispersion. Different colors of light will focus at different points, so an image of an object illuminated by white light will appear blurred.

Question 58

Define:

spherical aberration

Answer 58

Spherical aberration is the characteristic blurriness of an image created by a real spherical lens.

Spherical optics focus light rays well, as long as the light rays are near the center of the optic. Light rays closer to the optic’s edges, however, will not focus at exactly the same point, so images including light rays from those edges will appear blurred.

Question 59

How can the overall magnification of a multiple-lens system be calculated?

Answer 59

The overall magnification is simply the product of the magnifications of all of the individual lenses.

Question 60

The object of a two-lens optical system is 15 cm tall. If the magnifications of the lenses are m₁ = +2 and m₂ = +1/3, respectively, what is the overall height of the final image?

Answer 60

The final image is 10 cm tall.

The overall magnification of the system is the product of the two magnifications:

m_tot = m₁m₂ = (2)(1/3) = 2/3

The final image is upright and 2/3 the height of the original object, or 10 cm tall.

Question 61

If a microscope has a 10x eyepiece and a 40x objective, how large will a 10 µm object appear?

Answer 61

The object will appear to measure 4 mm.

The overall magnification is (10) (40) = 400x, so the final image size is:

10 x 10^-6 * 400 = 4 x 10^-3 m