Question

part h: application

1. a forensics lab technician is trying to solve an atm robbery. they have a picture of the suspect from the bank camera, but even when the photo is enlarged, the face is too blurry to use for identification. explain why the image is blurry when enlarged.

1. if a field of view is 5.0 mm at 60×, what is the field of view at 25×?

1. if you are using your microscope, and a specimen takes up 1/3 of the field of view at 1000×, what is its diameter?

a student viewed the letter “e” on a slide. the student could clearly see the letter when using the 4× and 10× objectives, but when the 40× objective was used, the “e” was no longer visible. give several possible reasons for this result.

Explanation:

Question 1

The image from the bank camera is a digital (or analog) image composed of pixels (or a grid of information). When an image is enlarged, if it's a low - resolution image, the number of pixels (or the amount of detail - carrying units) is fixed. Enlarging the image stretches these pixels (or the information grid), and there is no additional detail to fill in the enlarged space. So the image appears blurry as the individual pixels become more visible and the fine details are not present to maintain a clear image.

Step 1: Recall the relationship between field of view (FOV) and magnification (M)

The product of the field of view and the magnification is constant (assuming the lens system and other factors are such that the relationship \( \text{FOV}_1\times M_1=\text{FOV}_2\times M_2 \) holds). Let \( \text{FOV}_1 = 5.0\space\text{mm} \), \( M_1=60\times \), \( M_2 = 25\times \) and we need to find \( \text{FOV}_2 \).

Step 2: Rearrange the formula to solve for \( \text{FOV}_2 \)

From \( \text{FOV}_1\times M_1=\text{FOV}_2\times M_2 \), we can get \( \text{FOV}_2=\frac{\text{FOV}_1\times M_1}{M_2} \)

Step 3: Substitute the values into the formula

Substitute \( \text{FOV}_1 = 5.0\space\text{mm} \), \( M_1 = 60 \), and \( M_2=25 \) into the formula: \( \text{FOV}_2=\frac{5.0\space\text{mm}\times60}{25} \)
First, calculate \( 5.0\times60 = 300 \), then divide by 25: \( \frac{300}{25}=12\space\text{mm} \)

1. Focus Issue: The microscope might not be properly focused when using the \( 40\times \) objective. The \( 40\times \) objective has a shorter working distance and a higher magnification, so small changes in focus are more critical.
2. Parfocality Problem: The microscope may not be parfocal (the ability of a microscope to keep the specimen in focus when changing objectives). If it is not parfocal, when switching to the \( 40\times \) objective, the specimen may be out of the focal plane.
3. Specimen Position: The letter "e" may have moved out of the center of the field of view when switching objectives. The field of view is smaller at higher magnifications (\( 40\times \) has a smaller field of view than \( 4\times \) and \( 10\times \)), so if the specimen was not centered, it may be outside the field of view at \( 40\times \).
4. Depth of Field: The depth of field is smaller at higher magnifications. If the "e" is not in the same plane (e.g., the slide has some thickness or the "e" is not flat on the slide), at \( 40\times \) the part of the "e" that is in focus may not be the part that was visible at lower magnifications.
5. Lens Obstruction or Damage: The \( 40\times \) objective lens may be dirty, scratched, or obstructed, preventing a clear view of the specimen.

Answer:

The image is blurry when enlarged because it has a limited resolution (composed of a fixed number of pixels or detail - carrying units). Enlarging it stretches these units without adding new detail, making the image lose clarity as the individual pixels (or detail elements) become more noticeable.

Question 2