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θ = 1 . 22 λ D = x d , size 12{θ=1 "." "22" { {λ} over {D} } = { {x} over {d} } } {}

where d size 12{d} {} is the distance between the specimen and the objective lens, and we have used the small angle approximation (i.e., we have assumed that x size 12{x} {} is much smaller than d size 12{d} {} ), so that tan θ sin θ θ size 12{"tan"θ approx "sin"θ approx θ} {} .

Therefore, the resolving power is

x = 1 . 22 λd D . size 12{x=1 "." "22" { {λd} over {D} } } {}

Another way to look at this is by re-examining the concept of Numerical Aperture ( NA size 12{ ital "NA"} {} ) discussed in Microscopes . There, NA size 12{ ital "NA"} {} is a measure of the maximum acceptance angle at which the fiber will take light and still contain it within the fiber. [link] (b) shows a lens and an object at point P. The NA size 12{ ital "NA"} {} here is a measure of the ability of the lens to gather light and resolve fine detail. The angle subtended by the lens at its focus is defined to be θ = size 12{θ=2α} {} . From the figure and again using the small angle approximation, we can write

sin α = D / 2 d = D 2 d . size 12{"sin"α= { { {D} slash {2} } over {d} } = { {D} over {2d} } } {}

The NA for a lens is NA = n sin α size 12{ ital "NA"=n`"sin"α} {} , where n size 12{n} {} is the index of refraction of the medium between the objective lens and the object at point P.

From this definition for NA size 12{ ital "NA"} {} , we can see that

x = 1 . 22 λd D = 1 . 22 λ 2 sin α = 0.61 λn NA . size 12{x=1 "." "22" { {λd} over {D} } =1 "." "22" { {λ} over {2"sin"α} } =0 "." "61" { {λn} over { ital "NA"} } } {}

In a microscope, NA size 12{ ital "NA"} {} is important because it relates to the resolving power of a lens. A lens with a large NA size 12{ ital "NA"} {} will be able to resolve finer details. Lenses with larger NA size 12{ ital "NA"} {} will also be able to collect more light and so give a brighter image. Another way to describe this situation is that the larger the NA size 12{ ital "NA"} {} , the larger the cone of light that can be brought into the lens, and so more of the diffraction modes will be collected. Thus the microscope has more information to form a clear image, and so its resolving power will be higher.

Part a of the figure shows two small objects arranged vertically a distance x one above the other on the left side of the schematic. On the right side, at a distance lowercase d from the two objects, is a vertical oval shape that represents a convex lens. The middle of the lens is on the horizontal bisector between the two points on the left. Two rays, one from each object on the left, leave the objects and pass through the center of the lens. The distance d is significantly longer than the distance x. Part b of the figure shows a horizontal oval representing a convex lens labeled microscope objective that is a distance lowercase d above a flat surface. The oval’s long axis is of length capital D. A point P is labeled on the plane directly below the center of the lens, and two rays leave this point. One ray extends to the left edge of the lens and the other ray extends to the right edge of the lens. The angle between these rays is labeled acceptance angle theta, and the half angle is labeled alpha. The distance lowercase d is longer than the distance capital D.
(a) Two points separated by at distance x size 12{x} {} and a positioned a distance d size 12{d} {} away from the objective. (credit: Infopro, Wikimedia Commons) (b) Terms and symbols used in discussion of resolving power for a lens and an object at point P. (credit: Infopro, Wikimedia Commons)

One of the consequences of diffraction is that the focal point of a beam has a finite width and intensity distribution. Consider focusing when only considering geometric optics, shown in [link] (a). The focal point is infinitely small with a huge intensity and the capacity to incinerate most samples irrespective of the NA size 12{ ital "NA"} {} of the objective lens. For wave optics, due to diffraction, the focal point spreads to become a focal spot (see [link] (b)) with the size of the spot decreasing with increasing NA size 12{ ital "NA"} {} . Consequently, the intensity in the focal spot increases with increasing NA size 12{ ital "NA"} {} . The higher the NA size 12{ ital "NA"} {} , the greater the chances of photodegrading the specimen. However, the spot never becomes a true point.

The first schematic is labeled geometric optics focus. It shows an edge-on view of a thin lens that is vertical. The lens is represented by a thin ellipse. Two parallel horizontal rays impinge upon the lens from the left. One ray goes through the upper edge of the lens and is deviated downward at about a thirty degree angle below the horizontal. The other ray goes through the lower edge of the lens and is deviated upward at about a thirty degree angle above the horizontal. These two rays cross a point that is labeled focal point. The second schematic is labeled wave optics focus. It is similar to the first schematic, except that the rays do not quite cross at the focal point. Instead, they diverge away from each other at the same angle as they approached each other. The region of closest approach for the lines is called the focal region.
(a) In geometric optics, the focus is a point, but it is not physically possible to produce such a point because it implies infinite intensity. (b) In wave optics, the focus is an extended region.

Section summary

  • Diffraction limits resolution.
  • For a circular aperture, lens, or mirror, the Rayleigh criterion states that two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other.
  • This occurs for two point objects separated by the angle θ = 1 . 22 λ D size 12{θ=1 "." "22" { {λ} over {D} } } {} , where λ size 12{λ} {} is the wavelength of light (or other electromagnetic radiation) and D size 12{D} {} is the diameter of the aperture, lens, mirror, etc. This equation also gives the angular spreading of a source of light having a diameter D size 12{D} {} .

Questions & Answers

Give an example (but not one from the text) of a device used to measure time and identify what change in that device indicates a change in time.
David Reply
hour glass, pendulum clock, atomic clock?
S.M
tnks
David
how did they solve for "t" after getting 67.6=.5(Voy + 0)t
Martin Reply
Find the following for path D in [link] : (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.
David Reply
the topic is kinematics
David
can i get notes of solid state physics
Lohitha
just check the chpt. 13 kinetic theory of matter it's there
David
is acceleration a fundamental unit.
David Reply
no it is derived
Abdul
no
Nisha
K thanks
David
no it's not its derived
Emmanuel
hi
Gift
Hello
Gift
hello gift
Emmanuel
hello
David
Hello Emmanuel
Gift
how are you gift
Emmanuel
I'm good
Gift
that's good
Emmanuel
how are you too
Gift
am cool
Emmanuel
spending time summarizing
Emmanuel
broadening my horizon
Emmanuel
I am fin
Longwar
ok
Gift
hi guys can you teach me how to solve a logarithm?
Villaflor Reply
how about a conceptual framework can you simplify for me? needed please
Villaflor
Hello what happens when electrone stops its rotation around its nucleus if it possible how
Afzal
I think they are constantly moving
Villaflor
yep what is problem you are stuck into context?
S.M
not possible to fix electron position in space,
S.M
Physics
Beatriz
yes of course Villa flor
David
equations of kinematics for constant acceleration
Sagcurse Reply
A bottle full of water weighs 45g when full of mercury,it weighs 360g.if the empty bottle weighs 20g.calculate the relative density of mercury and the density of mercury....pls I need help
Lila Reply
well You know the density of water is 1000kg/m^3.And formula for density is density=mass/volume Then we must calculate volume of bottle and mass of mercury: Volume of bottle is (45-20)/1000000=1/40000 mass of mercury is:(360-20)/1000 kg density of mercury:(340/1000):1/50000=(340•40000):1000=13600
Sobirjon
the latter is true
Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture...take density of water as 1g/cm3 and density of liquid 1.2g/cm3
Lila
plz hu can explain Heisenberg's uncertainty principle
Emmanuel Reply
who can help me with my problem about acceleration?
Vann Reply
ok
Nicholas
how to solve this... a car is heading north then smoothly made a westward turn during the travel the speed of the car remains constant at 1.5km/h what is the acceleration of the car? the total travel time of the car as it smoothly changed its direction is 15 minutes
Vann
i think the acceleration is 0 since the car does not change its speed unless there are other conditions
Ben
yes I have to agree, the key phrase is, "the speed of the car remains constant...," all other information is not needed to conclude that acceleration remains at 0 during the entire time
Luis
who can help me with a relative density question
Lila
1cm3 sample of tin lead alloy has mass 8.5g.the relative density of tin is 7.3 and that of lead is 11.3.calculate the percentage by weight of tin in the alloy. assuming that there is no change of volume when the metals formed the alloy
Lila
morning, what will happen to the volume of an ice block when heat is added from -200°c to 0°c... Will it volume increase or decrease?
adefenwa Reply
no
Emmanuel
hi what is physical education?
Kate
BPED..is my course.
Kate
No
Emmanuel
I think it is neither decreases nor increases ,it remains in the same volume because of its crystal structure
Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture. take density of water as 1g/cm3 and density of liquid as 1.2g/cm3
Lila
Sorry what does it means"no changes in volume occured"?
Sobirjon
volume can be the amount of space occupied by an object. But when an object does not change in shape it will still occupy the same space. Thats why the volume will still remain the same
Ben
Most soilds expand when heated but if it changes state at 0C it will have less volume. Ice floats because it is less dense ie a larger mass per unit volume.
Richard
how to calculate velocity
Okwethu Reply
v=d/t
Emeka
his about the speed?
Villaflor
how about speed
Villaflor
v=d/t
Nisha
hello bro hw is life with you
Jacob Reply
Mine is good. How about you?
Chase
Hi room of engineers
lawan Reply
yes,hi sir
Okwethu
hello
akinmeji
Hello
Mishael
hello
Jerry
hi
Sakhi
hi
H.C
so, what is going on here
akinmeji
u are all wlc just ask your question anybody. can answer
Ajayi
good morning ppl
ABDUL
If someone has not studied Mathematics enough yet, should theu study it first then study Phusics or Study Basics of Physics whilst srudying Math as well?
Riaz Reply
whether u studied maths or not, it is advisable to start from d basics cuz it is essential to know dem
Nuru
yea you are right
Badmus
wow, you got this w/o knowing math
Thomas
I guess that's it
Thomas
later people
Thomas
mathematics is everywhere
Anand
thanks but dat doesn't mean it is good without maths @Riaz....... Maths is essential in sciences particularly wen it comes to PHYSICS but PHYSICS must be started from the basic which may also help in ur mathematical ability
Nuru
A hydrometer of mass 0.15kg and uniform cross sectional area of 0.0025m2 displaced in water of density 1000kg/m3.what depth will the hydrometer sink
Lila
16.66 meters?
Darshik
16.71m2
aways
,i have a question of let me give answer
aways
the mass is stretched a distance of 8cm and held what is the potential energy? quick answer
aways
oscillation is a to and fro movement, it can also be referred to as vibration. e.g loaded string, loaded test tube or an hinged door
Olatunji Reply
Practice Key Terms 1

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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