Biology, 06.07.2019 18:30, iris7324

1. what is the total magnification of an object if the ocular lens magnification is 20x and the objective lens magnification is 45x? 2. which objective lens is in place if the object you are viewing is magnified 1000x assuming an ocular lens magnification of 10x? 3. what is the diameter of the field of view (dfv) of a 1000x objective lens if the dfv of a 400x objective lens is 500 μ? express your answer in mm. 4. what is the dfv of a 40x objective lens if the dfv of a 10x objective lens is 3 mm? express your answer in μ. 5. when viewing an organism using the 40x objective lens from question 4, you estimate 6 organisms could fit across the dfv if they were laid end-to-end and 20 could fit is stacked side by-side. what is the length and width of this organism (in microns)? 6. what is the dfv of a 25x objective lens if the dfv of a 100x objective lens is 1.5 mm? 7. using the 100x objective lens from question 6, you

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answered: melkumathurin

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 μDFV of the 40× objective is 12 mmLength 6 mm, width = 0.6 mmThe DFV of the 25× objective is 6 mmThe 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×

$2. \ Objective \ lens \ magnification = \mathbf{\dfrac{Total \ magnification}{Eye \ piece \ magnification}}$

Therefore;

$Objective \ lens \ magnification = \dfrac{1000 \times }{10\times } = 100 \times$

3. We have;

m₁ × DFV₁ = m₂ × DFV₂

$DFV_{1000\times } = \mathbf{\displaystyle \frac{500 \, \mu \times 400 \times}{1000 \times }} = 200 \, \mu$

DFV of a 1000× objective lens = 200 μ

4. The DFV of the 10 objective lens is found as follows;

$DFV_{40\times } = \displaystyle \frac{3 \, mm \times 40 \times}{10 \times } = \underline{ 12 \, mm}$

5. The length and width of the organisms are found as follows;

$Length = \dfrac{12 \, mm}{2} = \underline{ 6 \, mm}$

$Width = \dfrac{12 \, mm}{20} = \underline{ 0.6 \, mm}$

6. The DFV of the length is found as follows;

$DFV_{25\times } = \displaystyle \frac{1.5 \, mm \times 100 \times}{25 \times } = \underline{ 6 \, mm}$

7. The organism's length and width is found as follows;

$\displaystyle Length = \dfrac{6 \, mm}{12} = \frac{1}{2} \, mm =\underline{ 0.5 \, mm}$

$\displaystyle Width = \dfrac{6 \, mm}{30} = \underline{ 0.2 \, mm}$

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answered: terrellsmith84

1. 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. 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. 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. 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. 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. 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)?

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

answered: Guest

answer; the above statement is (true);

answered: Guest

one centimetre is equal to 0.01 metres