Fluorescence Microscopy and Bioimage Processing

Question 1

equation for magnification

Answer 1

v/u=magnification
v = image distance
u = object distance
*the closer an object is to the lens, the higher the magnification

Question 2

equation for total magnification

Answer 2

magnification eyepiece x magnification objective = total magnification

Question 3

what is an inverted microscope?

Answer 3

• the object being viewed is upside down
• the observer looks up into the microscope from below
• used for observing samples suspended in liquid

Question 4

what is an upright microscope?

Answer 4

• object being viewed is in an upright position
• observer looks down into microscope from above
• used for observing samples on a slide

Question 5

what does the eye piece do/consist of?

Answer 5

ocular and tube,
magnifies the primary image produced by the objective

Question 6

what is the role of the condenser?

Answer 6

aligns light rays into straight path. Adjust it for the objective

Question 7

what is spherical aberration?

Answer 7

a phenomenon in which the image can be blurry because the light rays are not focused on the same point.

Question 8

what is chromatic aberration?

Answer 8

a phenomenon where the image is sharp, but different colours show.

Question 9

definition of numerical aperture

Answer 9

The numerical aperture (NA) of a microscope objective is a measure of its ability to gather light and
resolve fine specimen detail at a fixed object distance

Question 10

calculate numerical aperture

Answer 10

Numerical Aperture (NA) = n x (sin m(mew)) ​

m = angle of one-half the angular aperture (A)​

n = Refractive Index of imaging medium

Question 11

relationship between numerical aperture and resolution?

Answer 11

The higher the total numerical aperture, the better the resolution.

Question 12

refraction

Answer 12

bending of light, occurs as light passes from one medium into another medium with a
different refractive index.

Question 13

refractive index

Answer 13

a dimensionless number that gives the indication of the light bending ability of that medium.

Question 14

resolution

Answer 14

measure of the microscopes ability to distinguished the smallest distance between two points
on a specimen as separate entities.

Question 15

resolution calculation

Answer 15

R = λ/(2 x NA) (where λ is the wavelength and NA is the numerical aperture)

Question 16

what is fluorescence?

Answer 16

Fluorescence is the property of some atoms/molecules to absorb light (the excitation: Ex) of short
wavelength and emitting (Em) light of longer wavelength.

Question 17

what is the Stokes shift?

Answer 17

The distance between the excitation and emission peaks is known as the Stokes shift.

Question 18

give an advantage to fluorescence microscopy

Answer 18

collect images in more than one colour

Question 19

what is confocal microscopy?

Answer 19

Confocal microscopy is an advanced light microscopy method which utilises a ‘pinhole’ to eliminate
out of focus light.

Question 20

what are the stages of image processing?

Answer 20

1) image acquisition, 2) image processing and 3) image analysis.​

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Question 21

what filters are used in image processing?

Answer 21

deconvolution, Gaussian blur, subtract background

Question 22

what is the purpose of filters?

Answer 22

improve the effectiveness of image analysis

Question 23

what is deconvolution?

Answer 23

corrects the systematic error of blur (loss of contrast in smaller features) and reconstructs true image.​

Question 24

what is Gaussian blur?

Answer 24

image is convoluted with a Gaussian function for smoothing to reduce the image noise.

Question 25

what is subtract background?

Answer 25

Removes backgrounds from images. A local background value is determined for every pixel by averaging over a very large area.​

Question 26

what is thresholding?

Answer 26

pixels are divided into foreground (bright structures) and background (dark structures).

Question 27

what is segmentation?

Answer 27

• Image segmentation is the process of partitioning a digital image into multiple segments.
• reduces the complexity of the image and makes subsequent analysing easier
• erode and dilate