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**Magnification **

Traditionally, three types of magnification are discussed: relative distance magnification, relative size magnification, and angular magnification.

**a. Relative Distance Magnification**

The easiest way to magnify an object is to bring the object closer to the eye. By moving the object of regard closer to the eye, the size of the image on the retina is enlarged. Children with visual impairments do this naturally. Adults will require reading glasses to have the object in focus.

- Relative Distance Magnification = r/d where r = reference or original working distance and d = new working distance
- Example
- Original working distance = 40cm
- New working distance = 10cm
- Relative Distance Magnification (RDM) = 40/10 = 4x

With reading glasses, as the lens power increases, the working distance decreases. The reading glasses do not magnify by their power alone when worn in the spectacle plane. Magnification occurs because the lens strength requires the individual using them to hold things closer to have the object in focus.

*b. Relative Size Magnification *

Relative size magnification enlarges the object while maintaining the same working distance, for instance, as observed with large print.

- Relative Size Magnification = S2/S1 where S1 = original size and S2 = the new size
- Example
- Original size = 1M
- New size = 2M
- Relative Size Magnification (RSM) = 2/1 = 2x

*c. Angular Magnification: *(Figure 34)

Angular magnification occurs when the object is not changed in position or size, but has an optical system interposed between the object and the eye to make the object appear larger.

Examples: Telescopes and hand magnifiers

Click on image to enlarge.

This optical system produces a virtual image smaller than the original object but much closer to the eye. The image has a larger angular subtense than the original object; therefore, the objects appear larger when seen through this optical system even though the virtual image is smaller than the object.

Angular magnification is the ratio of the angular subtense of the image produced by a device divided by the angular subtense of the original object. Angular magnification takes into account not only the size of an image, but also its distance from the observer.

*d. Magnification Basics*

- Perceived size is proportional to the size of the object’s image on the retina.
- Retinal image size is proportional to the object’s angular subtense.
- Angular subtense is directly proportional to the object size and inversely proportional to the object’s distance from the observer.

Magnification looks at the ratio of object size (Y) to the image size (Y’) or the ratio of the angular subtense of the image viewed with the optical system to the angular subtense of the object viewed without the optical system.

- Plus (+) indicates the image is upright
- Minus (-) indicates the image is inverted.
- When the image is smaller than the object, magnification is numerically between 0-1.
- If the absolute number is greater than 1, the image is larger.
- If the absolute number is equal to 1, it is the same size.
- A magnification of –4 implies that the image is inverted and 4x larger than the object
- A magnification of 0.2 implies that the image is erect and 1/5 the size of the object.
- If the object and image are on the same side of the lens, the image is erect, if not, the image is inverted.
- Generally, if the image is located farther from the lens than the object is, the image is larger than the object, if the image is closer to the lens than the object; the image is smaller than the object.

*e. Transverse/Linear Magnification*(Figure 35)The ratio of the image size to the object size or image vergence to object vergence is called transverse or linear magnification. M T = I/O=U/V = v/u

Click on image to enlarge.

**Question:** An object is placed 20cm in front of a +10.00 diopter lens. What will the resultant linear magnification be?

**Answer:** -1

**Explanation:** In order to calculate linear magnification for a single lens system, one must know only the object distance and the image distance and/or the object vergence and the image vergence.

The formula is *Magnification = image distance (v)/object distance (u) = U/V*.

If the object distance is 20cm, the rays incident on the lens have a vergence of 100/-20 = -5.00D. After refraction, through the +10.00 diopter lens, the rays have a vergence of

-5.00 + (+10.00D) = +5.00D. Therefore, a real image is formed 100/+5.00D = +20cm behind the lens.

As it turns out, the object distance of 20cm and the image distance of 20cm, are equal, so the magnification is -1. Also the object vergence is –5.00D and the image vergence is +5.00D giving a magnification of –5.00/+5.00 of –1, indicating the image is inverted.

**Question: **Consider an optical system consisting of two lenses in air. (Figure 36) The first lens is +5.00D, the second lens is +8.00D and they are separated by 45 cm. If an object is 1 meter in front of the first lens, where is the final image and what is the transverse magnification?

**Answer:** To analyze a combination of lenses, we must look at each lens individually. The thin lens equation (U + D = V = 1/-1 + (+5.00) = -1 +5 = +4.00D and 100/+4.00 = +25cm) shows that the first lens produces an image 25cm behind itself, with the magnification (M = U/V = -1/+4 = –0.25). Light converges to the image and then diverges again. The image formed by the first lens becomes an object for the second lens. The image is 20 cm in front of the second lens, thus light strikes the second lens with a vergence of (1/-0.20) –5.00D and forms an image 33cm behind the second lens (-5 + (+8) = +3.00, 100/+3.00 = +33cm). Transverse magnification for the second lens alone is –5.00D/3.00D or –1.66. The total magnification is the product of the individual magnification –1.66 x –0.25 = 0.42.

Click on image to enlarge.

*f. Axial Magnification = M 1 X M 2: *(Figure 37)

Axial magnification is used when talking about objects that do not occupy a single plane (3D objects). Axial magnification is the distance, along the optical axis, between the two image planes divided by the distance between the two object planes (extreme anterior and posterior points on the object with their conjugate image points). Axial magnification is proportional to the product of the transverse magnifications for the pair of conjugate planes at the front and back of the object.

Click on image to enlarge.

For objects with axial dimensions that are relatively small, M 1 and M 2 are usually very close in numerical value, which leads to the approximate formula of:

Axial Magnification = M 2

Where M is the transverse magnification for any pair of the object’s conjugates.

**Question:** The front of a 5cm thick object is 20cm in front of a +9.00D lens. Calculate the axial magnification using both formulas.** **

**Answer:** The two faces of the object are positioned 20 and 25cm in front of the lens. From the vergence equation, the face located 20cm (U = 100/-20 = -5.00D) in front of the lens is imaged 25 cm behind the lens (U + D = V = -5 + (+9) = +4D, v = 100/+4 = +25cm). The magnification = image distance/object distance = +25/-20 = -1.25X.

The other side of the object located 25cm (100/-25 = -4.00D) from the lens is imaged at 20cm (U + D = V = -4 + (+9) = +5D, v = 100/+5 = +20cm) behind the lens with a magnification of 20/-25 = -0.8

Using the approximation formula, the axial magnification is either (-1.25) 2 = 1.56 or (-0.8) 2 = 0.64, depending on which plane we choose.

Using the exact formula, the axial magnification is -0.8 x -1.25 = 1.00

**Question:** An example of the importance of axial magnification is the evaluation of optic nerve cupping using indirect ophthalmoscopy. The cup can be evaluated using a +20.00D lens, but a +14.00D lens markedly improves the evaluation. What is the axial magnification of a 20D versus a 14D-condensing lens?

**Answer:** Lateral magnification produced through the indirect ophthalmoscope is the ratio of the total refracting power of the eye (60D) to the power of the condensing lens. The 14.00 diopter lens gives a slightly larger transverse magnification (60/20 = 3X versus 60/14 = 4.286X), but a significantly larger axial magnification because axial magnification increases as the square of transverse magnification (3X 2 = 9X versus 4.286X 2 = 18.37X). Larger axial magnification increases the distance between the optic nerve rim and the base of the cup in the aerial image, improving assessment of the cup.

*g. Effective Magnification = M e = dF*

Where d = reference distance in meters to the object (image is formed at infinity)

If d = 25cm than M e = F/4

If d = 40cm than M e = F/2.5

**Question:** A +24.00D lens is used as a hand held magnifier with the patient viewing an object that is 50cm from the eye and at the focal point of the lens. How much larger do things appear to the patient?

**Answer:** d = 0.50m, F= +24.00D, M e = dF = 0.50(24) = 12X

This indicates that closer working distances result in less effective magnification.

*h. Rated Magnification = Mr = F/4 *

Assumes that the individual can accommodate up to 4.00 diopters when doing close work which gives d = 25cm (25cm is the standard reference distance used when talking about magnification).

**Question:** A simple lens magnifier to be used as a low vision device is marked 5X (reference plane at 25cm). What would you expect to find when you measure the lens on a lensometer?

**Answer:** M = F/4 = 5 = F/4, F = 20D

**Question:** A view of the retina is obtained through an indirect ophthalmoscope, using a 30-diopter lens. The observer is 40cm from the arial image. What is the perceived lateral magnification?

**Answer:** 1.25x

**Explanation**: Lateral magnification produced through the indirect ophthalmoscope is the ratio of the total refracting power of the eye (60D) to the power of the condensing lens (30D), assuming the standard reference distance for magnification of 25cm from the observer to the arial image. If the distance is greater than 25cm, the lateral magnification is multiplied by the ratio of the standard reference distance, 25cm, to the distance in question, 40cm.

60/30D = 2x magnification at 25cm (2 x 25cm/40cm) = 1.25x magnification

*i. Conventional Magnification = Mc = dF + 1 *

The underlying assumption in this equation is that the patient is “supplying” one unit (1X) of magnification

**Question:** Which patient needs more magnification and which patient needs the stronger lens? Patient A wants to read 1M print and has a near acuity of 2M using a +5.00 diopter add at 20cm. Patient B also wants to read 1M and has an acuity of 3M with a +2.50 diopter add at 40cm.

**Answer:** Patient A reads 2M print and wants to read 1M print, therefore, 2M/1M = 2x magnification. F s needed is +5D X 2 = +10.00 diopters.

Patients B needs 3M/1M or 3x magnification and has F s of +2.5x3 = +7.5D. Even though Patient B needs 1 ½ times the amount of magnification Patient A does, (3M versus 2M to start) he actually requires a weaker lens than Patient A does.

This apparent paradox in magnification is because we are comparing apples to oranges when we use different distances. To compensate for different viewing distances, change patient B’s working distance to 20cm, the same as patient A. He would then see 1.5M print using a 5.00 diopter add for the 20cm working distance. 1.5M/1M = 1.5x times 5 diopters which = 7.5 diopters of magnification needed.

*j. Magnification Ratings*

Some companies use F/4 (Rated Magnification) while others use (F/4) + 1 (Conventional Magnification) to determine magnification strength for their magnifiers. This is why dioptric power, which is an absolute value and is the same under all conditions, is a better way to discuss the magnification needs of an individual.

*k. Determining Needed Magnification *

- Magnification needs are based on the initial reference value and the desired final value. Clinically, it is the entrance acuity divided by the goal acuity (VA/VA’).

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