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Chapter 24: Problem 70

A convex lens of power +12 D is used as a magnifier to examine a wildflower. What is the angular magnification if the final image is at (a) infinity or (b) the near point of \(25 \mathrm{cm} ?\)

Short Answer

Expert verified
The angular magnification when the final image is at infinity is 3, and when the final image is at the near point (25 cm), it is 2.

Step by step solution

01

Calculate the focal length of the lens

We are given the power of the lens (P) and we can find the focal length (f) using the formula: \( f = \frac{1}{P} \) In our case, P = +12 D. So the focal length of the lens is: \( f = \frac{1}{12} = 0.0833 \mathrm{m} \)
02

Find the object distance (u) when the image is at infinity

When the final image is at infinity, the object is placed at the focal point of the lens (f). In this case, the object distance u is equal to the focal length (0.0833 m): \( u = f = 0.0833 \mathrm{m} \)
03

Calculate the angular magnification (a)

We can now find the angular magnification using the formula: \( M = \frac{d}{u} \) Here, d is the least distance of distinct vision, which is 25 cm or 0.25 m. So the angular magnification for case (a) is: \( M = \frac{0.25}{0.0833} = 3 \) Thus, the angular magnification when the final image is at infinity is 3.
04

Find the object distance (u) when the image is at the near point

When the image is at the near point (25 cm), we need to rearrange the lens formula and solve for u: \( \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \) We know the focal length (f) and the image distance (v) is equal to the near point distance, which is 25 cm or 0.25 m. Substituting these values, we get: \( \frac{1}{u} = \frac{1}{0.0833} - \frac{1}{0.25} = 8 \) Thus, the object distance (u) when the image is at the near point is: \( u = \frac{1}{8} = 0.125 \mathrm{m} \)
05

Calculate the angular magnification (b)

Now, we can find the angular magnification for case (b) using the formula: \( M = \frac{d}{u} \) Using the object distance (u) calculated in Step 4, we have: \( M = \frac{0.25}{0.125} = 2 \) Thus, the angular magnification when the final image is at the near point is 2. In conclusion, the angular magnification of the convex lens is 3 when the final image is at infinity and 2 when the final image is at the near point (25 cm).

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Most popular questions from this chapter

You plan to project an inverted image \(30.0 \mathrm{cm}\) to the right of an object. You have a diverging lens with focal length \(-4.00 \mathrm{cm}\) located \(6.00 \mathrm{cm}\) to the right of the object. Once you put a second lens at \(18.0 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high. what is the image height?
An insect that is 5.00 mm long is placed \(10.0 \mathrm{cm}\) from a converging lens with a focal length of \(12.0 \mathrm{cm} .\) (a) What is the position of the image? (b) What is the size of the image? (c) Is the image upright or inverted? (d) Is the image real or virtual? (e) What is the angular magnification if the lens is close to the eye?
Two lenses, separated by a distance of \(21.0 \mathrm{cm},\) are used in combination. The first lens has a focal length of \(+30.0 \mathrm{cm} ;\) the second has a focal length of \(-15.0 \mathrm{cm} .\) An object, $2.0 \mathrm{mm}\( long, is placed \)1.8 \mathrm{cm}$ before the first lens. (a) What are the intermediate and final image distances relative to the corresponding lenses? (b) What is the total magnification? (c) What is the height of the final image?
The distance from the lens system (cornea + lens) of a particular eye to the retina is \(1.75 \mathrm{cm} .\) What is the focal length of the lens system when the eye produces a clear image of an object \(25.0 \mathrm{cm}\) away?
Unless the problem states otherwise, assume that the distance from the comea- lens system to the retina is \(2.0 \mathrm{cm}\) and the normal near point is \(25 \mathrm{cm}.\) If the distance from the lens system (cornea + lens) to the retina is $2.00 \mathrm{cm},$ show that the focal length of the lens system must vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm}\) to see objects from $25.0 \mathrm{cm}$ to infinity.
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