Respuesta :
1. Answer: 900x
A typical microscope has two lens: ocular lens which located near the eye, and objective lens which located near the object. The image will undergo magnification of both lens. The total magnification of the object would be the product of both lens multiplication, not the sum. The multiplication will be: 20x * 45x= 900x
total mag= ocular * object
total mag= 20x * 45x
total mag= 900x
2. Answer: 100x
The total magnification of the object would be the product of both lens multiplication. If the total magnification 1000x, that mean it was the product of ocular lens and objective lens. If ocular lens magnification is 10x, objective lens magnification would be:
total mag= ocular * object
1000x= 10x*object
object=1000x/ 10x=
object=100x
3. Answer: 0.2 mm
The area of the microscope that can be viewed by the eye should be the same. But since the magnification is different, the actual area that it represent will be different. Microscope with bigger magnification will have smaller diameter of the field of view(DFV). Remember that micrometer(μm) is 1/1000 of millimeter(mm). The DFV would be:500 μm/ ( 1000x/400x)= 200μm= 0.2 mm
4. Answer: 750μm
This question is similar to number 3 question. Remember that 1 millimeter(mm) equal to 1000 micrometer(μm) .
The DFV of a 10x objective lens is 3 mm. Then the DFV of 40x lens would be:
DFV2= DFV1/ (mag1/mag2)
DFV2= 3mm/ ( 40x/10x)= 0.75 mm= 750μm
5. Answer: 125μm x 37.5μm
The lens DFV of microscope from question 4 is 750μm, so the width and length of the area would be 750μm.
If 6 organisms could fit across the DFV if they were laid end-to-end, the length would be: 750μm/ 6 organism= 125μm
If 20 organism could fit is stacked side by-side, then the width would be: 750μm/ 20 organism= 37.5μm
6. Answer: 6mm
This question is similar to number 3 and number 4 question. Remember that 1 millimeter(mm) equal to 1000 micrometer(μm).
The DFV of a 100x objective lens is 1.5 mm. Then the DFV of 25x lens would be:
DFV2= DFV1/ (mag1/mag2)
DFV2= 1.5mm/ (100x/25x)= 6 mm
7.Using the 100x objective lens from question 6, you estimate 12 organisms could fit across the DFV if they were laid end-to-end and 30 could fit is stacked side-by-side. What is the length and width of this organism (in microns)?
Answer: 0.5mm x 0.2mm or 500μm x 200μm
The lens DFV of microscope from question 6 is 6 mm, so the width and length of the area would be 6 mm.
If 12 organisms could fit across the DFV if they were laid end-to-end, the length would be: 6mm/ 12 organism=0.5mm
If 30 organism could fit is stacked side by-side, then the width would be: 6mm/ 30 organism= 0.2mm
The relationship between the magnification of a lens and its DSV is an
inverse relationship.
- The total magnification is 900×
- The objective lens magnification 100×
- DFV of a 1000× objective lens is 200 μ
- DFV of the 40× objective is 12 mm
- Length 6 mm, width = 0.6 mm
- The DFV of the 25× objective is 6 mm
- The organisms length is 0.5 mm, the width is 0.2 mm
Reasons:
1. Total magnification = Eye piece lens magnification × Objective lens magnification
Therefore;
The total magnification = 20× × 45× = 900×
[tex]2. \ Objective \ lens \ magnification = \mathbf{\dfrac{Total \ magnification}{Eye \ piece \ magnification}}[/tex]
Therefore;
[tex]Objective \ lens \ magnification = \dfrac{1000 \times }{10\times } = 100 \times[/tex]
3. We have;
m₁ × DFV₁ = m₂ × DFV₂
[tex]DFV_{1000\times } = \mathbf{\displaystyle \frac{500 \, \mu \times 400 \times}{1000 \times }} = 200 \, \mu[/tex]
DFV of a 1000× objective lens = 200 μ
4. The DFV of the 10 objective lens is found as follows;
[tex]DFV_{40\times } = \displaystyle \frac{3 \, mm \times 40 \times}{10 \times } = \underline{ 12 \, mm}[/tex]
5. The length and width of the organisms are found as follows;
[tex]Length = \dfrac{12 \, mm}{2} = \underline{ 6 \, mm}[/tex]
[tex]Width = \dfrac{12 \, mm}{20} = \underline{ 0.6 \, mm}[/tex]
6. The DFV of the length is found as follows;
[tex]DFV_{25\times } = \displaystyle \frac{1.5 \, mm \times 100 \times}{25 \times } = \underline{ 6 \, mm}[/tex]
7. The organism's length and width is found as follows;
[tex]\displaystyle Length = \dfrac{6 \, mm}{12} = \frac{1}{2} \, mm =\underline{ 0.5 \, mm}[/tex]
[tex]\displaystyle Width = \dfrac{6 \, mm}{30} = \underline{ 0.2 \, mm}[/tex]
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