Question

Mordern microscopes are more likely to use a camera than human viewing. This is accomplished by replacing the eyepiece in figure $35.14$with a photo-ocular that focuses the image of the objectives to a real image on the sensor of a digital camera. A typical sensor is $22.5mm$wide and consists of $56254.0\mu m$wide pixels. suppose a microscopist pairs a $40X$objectives with a $2.5X$photo-ocular

a. what is the field of view? That is what width on the microscope stage in $mm$fills the sensor?

b. The photo of a cell is $120pixels$in a diameter. what is the cell's actual diameter in$\mu m$?

Short answer

a. The field of view is ${\omega }_1=0.225mm$.

b. The cell's actual diameter is$d=4.8\mu m$.

Step-by-step solution

Part (a) Step 1: Given information 

We have given that:

Magnification of object$m_{obj}=40$,

width of typical sensor${\omega }_s=22.5mm$and

magnification of photo ocular$M_{\alpha }=2.5$.

We need to find the magnification M .

Part (a) Step 2: Simplification 

Let us find Magnification M:

$M=m_{obj}{M}_{\alpha }$ (substitute values in equation.)

$M=\left(40\right)(2.5)$

$M=100$

Therefore width ${\omega }_1$will be,

${\omega }_1=\frac{{\omega }_s}{M}$

substituting the values we get ,

${\omega }_1=0.225mm$

Here,${\omega }_1$is the field of view.

part (b) Step 1: Given Information 

We need to find the cell's actual diameter in $\mu m.$

part (b) step 2: Simplification

Firstly we have to find width ${\omega }_2$of the photo of the cell:

$$\begin{gathered} {\omega }_2=\left(120pixel\right)\left(4.0\mu m/pixel\right) \\ {\omega }_2=480\mu m \end{gathered}$$

Therefore Diameter is:

$$\begin{gathered} d=\frac{{\omega }_2}{M} \\ d=\frac{480\mu m}{100} \\ d=4.8\mu m. \end{gathered}$$ (substituting value in the equation.)

Here, $d$is the cell's actual diameter.