Optics Flashcards
How would you correct presbyopia?
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.
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.
True
True or False: Convex lenses cause rays to converge, while concave lenses cause them to diverge.
True
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
θincident ≥ θcritical
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?
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.
How do you calculate focal length with mirrors?
F= radius of curvature / 2
What is the thin lens equation?
1/f=1/o + 1/i
Howdo you calculate magnification?
m=- i/o
where m is magnification, i is image distance, and o is object distance
What is the focal point of a plane mirror?
f is infinite
For a plane mirror, what is the magnification?
Magnification for a plane mirror is equal to 1. The image is virtual and upright and not magnified.
True or False: Real images are always upright
False, real images are always inverted. Virtual images are always upright.
In a convex mirror, the image formed is always going to be…
a. real
b. virtual
c. it depends on the focal length
b. virtual
In a concave mirror, the image formed is always going to be…
a. real
b. virtual
c. it depends on the focal length
c. it depends on the focal length
In a concave mirror, if the object is less than the focal length, what is true about the image formed?
The image is upright and virtual
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?
The image is real and inverted
For a concave mirror, if the object is placed exactly on the focal length, what is true about the image formed?
There is no image formed when o=f
Convex mirrors ( converge or diverge ) light while convex lenses ( converge or diverge ) light
Convex mirrors diverge while convex lenses converge
Concave mirrors ( converge or diverge ) light while concave lenses ( converge or diverge ) light
Concave mirrors tend to converge while concave lenses tend to diverge light
With myopia, when does light converge?
Light converges before the retina.
How do you correct myopia?
Myopia, or nearsidedness, light converges before the retina as the lens refracts too much. You can correct this by using a diverging lens (concave).
How do you correct hyperopia?
What is a diopter?
1/m
What is the lensmaker’s equation?
1/f= (n-1) (1/r1 - 1/r2)
How do you calculate the power of a lens?
P= 1/ f
True or False: A shorter focal length means a more powerful lens
Yes, P=1/f
What is lens power measured in?
Diopters
Converging lens
Convex
Converging mirror
concave
Diversing lens
Concave
Diverging mirror
Convex
True/false: Negative image distance always implies a virtual image.
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.
For a given lens, object distance is 6/5 m and image distance is -3 m. The power of the lens is…
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.
True or false: A plane mirror only ever creates virtual images.
This statement is true. Light rays hitting a plane mirror will produce virtual images.
True or false: Myopia can be treated through the use of a lens with a negative focal length.
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.
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?
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.
A glass fiber carries a light digital signal long distances with a minimum loss of amplitude. What optical property of glass allows this phenomenon?
Reflection
Light can be carried along a distance within a transparent material by means of total internal reflection.
