Optics

Question 1

How would you correct presbyopia?

Answer 1

The problem with presbyopia (far-sightedness) is the image is formed behind the retina rather than on the retina. A converging lens will converge light rays before they can pass the retina and focus light closer to the retina.

Question 2

True or False: A ray of light that hits a plane mirror perpendicularly will be bounced back in the same direction, and rays of light that hit a plane mirror on an angle will reflect at the same but opposite angle to the normal.

Answer 2

True

Question 3

True or False: Convex lenses cause rays to converge, while concave lenses cause them to diverge.

Answer 3

True

Question 4

Light inside the thin glass tube of a laproscopic surgical device strikes the edge of the glass tube and is entirely reflected back into the tube, with none of the light exiting to the surrounding medium. Which of the following must be true? A.θincident = 90º B. θincident = 0º C.θincident ≥ θcritical ‘ D.θrefracted = θincident

Answer 4

θincident ≥ θcritical

Question 5

If a 10 cm tall object is placed 30 cm from a biconcave lens with a focal length of 10 cm, how tall will the image formed by the lens be?

Answer 5

2.5 cm

The size of the image is determined by the absolute value of magnification (m = -i/o). A biconcave lens is a diverging lens (f < 0) and we can use the second equation from the passage to solve for i.

1/-10 = 1/i + 1/30

1/i = 1/-10 – 1/30 = -3/30 – 1/30 = -4/30

i = -30/4 = -7.5

Now we can calculate m = -i/o = -(-7.5)/30 = 0.25, and its absolute value is 0.25. Thus, the image height will be ¼ of the object height, or (0.25)(10) = 2.5 cm.

Question 6

How do you calculate focal length with mirrors?

Answer 6

F= radius of curvature / 2

Question 7

What is the thin lens equation?

Answer 7

1/f=1/o + 1/i

Question 8

Howdo you calculate magnification?

Answer 8

m=- i/o

where m is magnification, i is image distance, and o is object distance

Question 9

What is the focal point of a plane mirror?

Answer 9

f is infinite

Question 10

For a plane mirror, what is the magnification?

Answer 10

Magnification for a plane mirror is equal to 1. The image is virtual and upright and not magnified.

Question 11

True or False: Real images are always upright

Answer 11

False, real images are always inverted. Virtual images are always upright.

Question 12

In a convex mirror, the image formed is always going to be…

a. real
b. virtual
c. it depends on the focal length

Answer 12

b. virtual

Question 13

In a concave mirror, the image formed is always going to be…

a. real
b. virtual
c. it depends on the focal length

Answer 13

c. it depends on the focal length

Question 14

In a concave mirror, if the object is less than the focal length, what is true about the image formed?

Answer 14

The image is upright and virtual

Question 15

In a concave mirror, if an object is placed further from the focal length such that o>f, what must be true about the image formed?

Answer 15

The image is real and inverted

Question 16

For a concave mirror, if the object is placed exactly on the focal length, what is true about the image formed?

Answer 16

There is no image formed when o=f

Question 17

Convex mirrors ( converge or diverge ) light while convex lenses ( converge or diverge ) light

Answer 17

Convex mirrors diverge while convex lenses converge

Question 18

Concave mirrors ( converge or diverge ) light while concave lenses ( converge or diverge ) light

Answer 18

Concave mirrors tend to converge while concave lenses tend to diverge light

Question 19

With myopia, when does light converge?

Answer 19

Light converges before the retina.

Question 20

How do you correct myopia?

Answer 20

Myopia, or nearsidedness, light converges before the retina as the lens refracts too much. You can correct this by using a diverging lens (concave).

Question 21

How do you correct hyperopia?

Question 22

What is a diopter?

Answer 22

1/m

Question 23

What is the lensmaker’s equation?

Answer 23

1/f= (n-1) (1/r1 - 1/r2)

Question 24

How do you calculate the power of a lens?

Answer 24

P= 1/ f

Question 25

True or False: A shorter focal length means a more powerful lens

Answer 25

Yes, P=1/f

Question 26

What is lens power measured in?

Answer 26

Diopters

Question 27

Converging lens

Answer 27

Convex

Question 28

Converging mirror

Answer 28

concave

Question 29

Diversing lens

Answer 29

Concave

Question 30

Diverging mirror

Answer 30

Convex

Question 31

True/false: Negative image distance always implies a virtual image.

Answer 31

This statement is true. An image produced by light rays reflecting off of a lens can be to the left or right side of a lens. By convention, images to the left of a lens are assigned a negative image distance. The only images produced to the left of a lens are the result of virtual image creation.

Question 32

For a given lens, object distance is 6/5 m and image distance is -3 m. The power of the lens is…

Answer 32

The power of a lens is equal to 1 divided by the focal length of the lens in meters. To find the focal length, we need to use the thin lens equation which can be written as:

1/i + 1/o = 1/f

Where f is the focal length, I is the image distance, and o is the object distance.

Substituting the values given in the problem into this equation, we find that:

1/(-3) + 1/(6/5) = 1/f

0.5 = 1/f

f = 2

Solving for f yields a focal length of 2 meters. Converting focal length to power yields a power of 0.5 diopters.

Question 33

True or false: A plane mirror only ever creates virtual images.

Answer 33

This statement is true. Light rays hitting a plane mirror will produce virtual images.

Question 34

True or false: Myopia can be treated through the use of a lens with a negative focal length.

Answer 34

This statement is true. A negative focal length is associated with a diverging lens. Myopia is a condition which occurs when a lens bends light too strongly and images are created in front of the retina. Myopia, or nearsightedness, is corrected by using a diverging lens. Diverging lenses cause light to diverge, which will counteract the greater convergence of light caused by a myopic lens. Therefore, myopia can be treated through the use of a lens with a negative focal length.

Question 35

A person’s eye has a lens to retina distance of 2 cm and an object is 1 meter away. What parts of the thin lens equation do these correspond to?

Answer 35

Recall the thin lens equation is 1/o + 1/i = 1/f.

The distance from the lens to retina is the image distance and the distance of the object to the lens is the object distance.

Question 36

A glass fiber carries a light digital signal long distances with a minimum loss of amplitude. What optical property of glass allows this phenomenon?

Answer 36

Reflection

Light can be carried along a distance within a transparent material by means of total internal reflection.