Bright-Field Microscopy of Transparent Objects 1117


Figure 1. In the absence of a specimen, a uniform illumination at the focal plane (2) of the objective (1) produces no contrast (a). The refractive specimen (3), which, for the sake of simplicity, is depicted here as a lens, alters the distribu- tion of intensity at the focal plane of the objective; the latter may lie inside (b) or outside (c) the sample. The intensity pattern at the focal plane is reproduced by the optical system and generates an image. In the case (b), one expects a lighter area in the central part, where the density of back-projected rays is higher. In the case (c), the area immediately outside the cone of light formed by the specimen is completely dark. (This effect can be easily observed by putting a lens under the microscope.)


usually true for live biological cells. Then, the angle of refrac- tion dα=α−β is small, and Equation (1) can be written as:


sinα sinðα - dαÞ


= sinα sinα - dsinα =


sinα - dαcosα = sinα


1 - dα = tanα =n: (2)


1


From here we find the refraction angle: dα = n - 1


ðÞtan α = n - 1


Figure 2. General setup in transmission brightfield imaging. 1,—slide; 2, focal plane; 3, object; 4, coverglass; 5, objective.


ðÞ


dhðxÞ dx ;


(3)


where h(x) is the height of the object boundary over the focal plane (it can be positive or negative), and x is the position on the focal plane where it would be intersected by the incident ray in the absence of the sample. Refraction causes a shift of the intersection point from x to x'. For small refraction angles dα:


x0 = x + hx ðÞdα = x + hx ðÞðÞ n - 1 ðÞdx;


dhðxÞ dx :


Image intensity I(x′) satisfies the equation: Ix0ðÞdx0 =I0 x


(4) (5)


where I0(x) is the incident light intensity. Equation (5) states the condition of energy conservation: all the incident rays that would arrive at the element dx without refraction are deflected to the element dx' in the presence of refraction. If we assume uniform illumination and use relative intensities, we can set I0(x)=1. Then:


Ix0ðÞ= dx dx0


= 1 dx0=dx ;


Figure 3. The diagram of rays crossing the sample. A vertical incident ray R impinges on the object boundary S at the angle α. Refraction causes its deflection from the vertical line by dα. The point of intersection of R and S lies at the distance h from the focal plane F. As a result, the point where the R ray crosses the focal plane F shifts from x to x′. The focal plane is designated as xy, and the refraction takes place in the xz-plane.


or, taking Equation (4) into account: Ix0ðÞ= 1 + n - 1


ðÞ hx d2hx = 1 + n - 1


"#-1 ðÞ


() ðÞ dx2 + dh x


 2


ðÞ dx


-1 ðÞdx hx


d ðÞ dx


dh x


-1 = 1 + n - 1 d2 h2 x


ðÞ 2


fg dx2


ðÞ : ðÞ ð7Þ (6)


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