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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

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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cm
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Can you compute that for me. Ty
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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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answer
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progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Who can show me the full solution in this problem?
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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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