OP - How Thin Lenses Work - Week 3

Question 1

Define a thin lens. Describe the transfer equation for thin lens.

Answer 1

When the axial thickness is small enough to be neglected, and d = 0.
Transfer equation becomes:
h2 = h1 = h
and so we end up with:
n'u' - nu = -h (F1 + F2)

Question 2

Bearing the thin lens equation in mind, describe the final thin lens equation form using the thin lens power formula.

Answer 2

Thin lens power formula:
F = F1 + F2
And so the final thin lens formula is:
n’u’ - nu = -h F

Question 3

Describe the formula for the power of a thin lens in terms of front and back surfaces and curvature.

Answer 3

F1 = (μ - n) C1 - power of front surface
F2 = (n' - μ) C2 - power of back surface
F1 + F2 becomes:
(C1 - C2)(μ - 1)
or
ΔC (μ - 1)

Question 4

Define the equation for curvature. Define the units of F, and conversion to D.

Answer 4

1/r
F units depends on the units of r, cm or m etc.
If taken in cm, /100 to convert to m, D is taken in m.

Question 5

Describe the sign conventions of curvature.

Answer 5

Consider the position of the lens.
Draw the required circles to form the lens.
If the middle of the circle is on the right of the lens, it is positive.
If on the left, it is negative.
Work from left to right when assigning C1 and C2.

Question 6

Consider two lens that are identical in shape and refractive index, but different signs (ie. mirror images). Are their powers the same?

Answer 6

Yes, because the formula for power F uses the change in curvature ΔC, so changing the signs has no effect on its power or the sign of the power.

Question 7

Define the lens equation.

Answer 7

n’/l’ - n/l = F

Question 8

Describe the lens equation with a point object at optical infinity, and the equation it leads to.

Answer 8

Becomes n’/l’ - n/-∞ = F
leading to:
l’ = n’/F

Question 9

Describe how to define the lens equation to form an image at optical infinity.

Answer 9

l' must be +∞
Lens equation becomes:
n'/∞ - n/l = F
Becoming:
l = -n / F

Question 10

Define the back and front focal points, and define the modified lens equation associated with both of them.

Answer 10

Back - collimated light focusing to a point, the back focal point
where l’ = n’/F
Front - a point source which is projected onto a thin lens, to form a collimated beam
where l = -n/F

Question 11

True or false:

n / F = n’ / F

Answer 11

False
In general, n / F ≠ n’ / F
However, if n = n’ = 1 (in air), it is true, this is the focal length, and f is used

Question 12

Describe what happens to n / F = n’ / F when n = n’ = 1.

Answer 12

1/F = 1/F
f is used, the focal length and so:
f = 1 / F

Question 13

Define focal length in terms of how it is used in ophthalmic optics.

Answer 13

f = 1 / F becomes F = 1 / f
or
power = 1 / focal length
Power is usually taken in D, and so focal length must be converted to m if in cm

Question 14

Define what the following means for power:
Long focal length
Short focal length

Answer 14

Long focal length - low power

Short focal length - high power

Question 15

Define whether the following parameters give real/virtual images/objects.
l' > 0
l' < 0
l < 0
l > 0

Answer 15

Images
l’ > 0 - real
l’ < 0 - virtual

Objects
l < 0 - real
i > 0 - virtual

Question 16

Define L’ and L, and how it relates to power. What are they called?

Answer 16

Vergence
L’ = n’ / l’ and L = n / l
F = L’ - L

Question 17

Consider a hyperbolic graph for positive power lens relating l and l’ as the x and y axes.
Which domains will give real and virtual images?
What are the vertical and horizontal asymptotes, and what do they represent?

Answer 17

Above the x axis (l’ > 0) - real image domain, below (l’ < 0) is the virtual image domain.
Horizontal asymptote = f
Vertical asymptote = -f
The asymptotes represent collimated beams.

Question 18

Consider a hyperbolic graph for negative power lens relating l and l’ as the x and y axes.
Which domains will give real and virtual images?
What are the vertical and horizontal asymptotes, and what do they represent?

Answer 18

Above the x axis (l’ > 0) - real image domain, below (l’ < 0) is the virtual image domain.
Horizontal asymptote = F (back focal point)
Vertical asymptote = F (front focal point)
The asymptotes represent collimated beams.