When One Uses a Magnifying Glass to Read Fine Print One Uses a

Geometric Eyes and Image Formation

16 The Simple Magnifier

Learning Objectives

By the end of this section, you will be able to:

Empathise the optics of a simple magnifier
Narrate the epitome created by a unproblematic magnifier

The apparent size of an object perceived past the eye depends on the angle the object subtends from the eye. Equally shown in (Effigy), the object at A subtends a larger angle from the centre than when it is position at indicate B. Thus, the object at A forms a larger image on the retina (run across $O{A}^{\prime }$ ) than when it is positioned at B (see $O{B}^{\prime }$ ). Thus, objects that subtend large angles from the heart appear larger because they form larger images on the retina.

Size perceived by an eye is determined by the angle subtended by the object. An image formed on the retina by an object at A is larger than an epitome formed on the retina past the same object positioned at B (compared prototype heights $O{A}^{\prime }$ to $O{B}^{\prime }$ ).

Two objects of the same size are shown in front of an eye. Object A is closer to the eye and forms an angle theta 2 with the optical axis. Object B is farther away and forms an angle theta 1 with the optical axis. Inside the eye, the rays strike the retina. Ray B prime is closer to the optical axis than ray A prime.

We accept seen that, when an object is placed inside a focal length of a convex lens, its paradigm is virtual, upright, and larger than the object (see role (b) of (Effigy)). Thus, when such an image produced by a convex lens serves as the object for the centre, as shown in (Effigy), the image on the retina is enlarged, because the prototype produced past the lens subtends a larger angle in the eye than does the object. A convex lens used for this purpose is chosen a magnifying glass or a elementary magnifier.

The simple magnifier is a convex lens used to produce an enlarged prototype of an object on the retina. (a) With no convex lens, the object subtends an angle ${\theta }_{\text{object}}$ from the heart. (b) With the convex lens in place, the image produced by the convex lens subtends an angle ${\theta }_{\text{image}}$ from the eye, with ${\theta }_{\text{image}}>{\theta }_{\text{object}}$ . Thus, the epitome on the retina is larger with the convex lens in place.

To account for the magnification of a magnifying lens, we compare the angle subtended past the image (created past the lens) with the angle subtended by the object (viewed with no lens), as shown in (Figure). We assume that the object is situated at the near indicate of the heart, because this is the object altitude at which the unaided eye tin can form the largest image on the retina. We will compare the magnified images created by a lens with this maximum image size for the unaided eye. The magnification of an epitome when observed by the eye is the angular magnification M, which is defined by the ratio of the angle ${\theta }_{\text{image}}$ subtended by the image to the angle ${\theta }_{\text{object}}$ subtended by the object:

$M=\frac{{\theta }_{\text{image}}}{{\theta }_{\text{object}}}.$

Consider the situation shown in (Figure). The magnifying lens is held a altitude $\mathcal{\ell }$ from the eye, and the image produced by the magnifier forms a distance 50 from the middle. We want to calculate the angular magnification for any arbitrary L and $\mathcal{\ell }$ . In the pocket-sized-angle approximation, the angular size ${\theta }_{\text{image}}$ of the epitome is ${h}_{\text{i}}\text{/}L$ . The angular size ${\theta }_{\text{object}}$ of the object at the nearly point is ${\theta }_{\text{object}}={h}_{\text{o}}\text{/}25\phantom{\rule{0.2em}{0ex}}\text{cm}$ . The athwart magnification is and then

$M=\frac{{\theta }_{\text{image}}}{{\theta }_{\text{object}}}=\frac{{h}_{\text{i}}\left(25\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}{L{h}_{\text{o}}}.$

Using (Effigy) for linear magnification

$m=\text{−}\frac{{d}_{\text{i}}}{{d}_{\text{o}}}=\frac{{h}_{\text{i}}}{{h}_{\text{o}}}$

and the thin-lens equation

$\frac{1}{{d}_{\text{o}}}+\frac{1}{{d}_{\text{i}}}=\frac{1}{f}$

in (Figure), we arrive at the following expression for the angular magnification of a magnifying lens:

$\begin{array}{cc}\hfill M& =\left(-\frac{{d}_{\text{i}}}{{d}_{\text{o}}}\right)\left(\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{L}\right)\hfill \\ & =\text{−}{d}_{\text{i}}\left(\frac{1}{f}-\frac{1}{{d}_{\text{i}}}\right)\left(\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{L}\right)\hfill \\ & =\left(1-\frac{{d}_{\text{i}}}{f}\right)\left(\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{L}\right)\hfill \end{array}$

From part (b) of the figure, we run into that the absolute value of the paradigm distance is $|{d}_{\text{i}}|=L-\mathcal{\ell }$ . Note that ${d}_{\text{i}}<0$ considering the epitome is virtual, so we tin dispense with the absolute value past explicitly inserting the minus sign: $-{d}_{\text{i}}=L-\mathcal{\ell }$ . Inserting this into (Figure) gives us the final equation for the angular magnification of a magnifying lens:

$M=\left(\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{L}\right)\left(1+\frac{L-\mathcal{\ell }}{f}\right).$

Note that all the quantities in this equation have to be expressed in centimeters. Oftentimes, we want the image to be at the near-point distance ( $L=25\phantom{\rule{0.2em}{0ex}}\text{cm}$ ) to go maximum magnification, and we hold the magnifying lens close to the eye ( $\mathcal{\ell }=0$ ). In this instance, (Effigy) gives

$M=1+\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{f}$

which shows that the greatest magnification occurs for the lens with the shortest focal length. In addition, when the epitome is at the near-point distance and the lens is held close to the center $\left(\mathcal{\ell }=0\right)$ , so $L={d}_{\text{i}}=25\phantom{\rule{0.2em}{0ex}}\text{cm}$ and (Figure) becomes

$M=\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=m$

where k is the linear magnification ((Effigy)) derived for spherical mirrors and thin lenses. Another useful situation is when the image is at infinity $\left(L=\text{∞}\right)$ . (Figure) then takes the course

$M\left(L=\text{∞}\right)=\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{f}.$

The resulting magnification is simply the ratio of the near-signal distance to the focal length of the magnifying lens, so a lens with a shorter focal length gives a stronger magnification. Although this magnification is smaller past 1 than the magnification obtained with the epitome at the near indicate, it provides for the almost comfortable viewing conditions, because the eye is relaxed when viewing a afar object.

By comparison (Figure) with (Figure), we see that the range of angular magnification of a given converging lens is

$\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{f}\le M\le 1+\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{f}.$

Magnifying a Diamond A jeweler wishes to inspect a 3.0-mm-bore diamond with a magnifier. The diamond is held at the jeweler'southward most point (25 cm), and the jeweler holds the magnifying lens shut to his eye.

(a) What should the focal length of the magnifying lens be to see a xv-mm-diameter prototype of the diamond?

(b) What should the focal length of the magnifying lens be to obtain $10\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$ magnification?

Strategy We demand to decide the requisite magnification of the magnifier. Because the jeweler holds the magnifying lens close to his eye, we tin can use (Figure) to observe the focal length of the magnifying lens.

Solution

The required linear magnification is the ratio of the desired image diameter to the diamond's actual diameter ((Figure)). Considering the jeweler holds the magnifying lens shut to his eye and the prototype forms at his about indicate, the linear magnification is the same every bit the athwart magnification, and so

$M=m=\frac{{h}_{\text{i}}}{{h}_{\text{o}}}=\frac{15\phantom{\rule{0.2em}{0ex}}\text{mm}}{3.0\phantom{\rule{0.2em}{0ex}}\text{mm}}=5.0.$

The focal length f of the magnifying lens may be calculated by solving (Figure) for f, which gives

$\begin{array}{cc}\hfill M& =1+\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{f}\hfill \\ \hfill f& =\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{M-1}=\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{5.0-1}=6.3\phantom{\rule{0.2em}{0ex}}\text{cm}\hfill \end{array}$
To get an image magnified past a factor of 10, we again solve (Effigy) for f, but this time we utilize $M=10$ . The issue is

$f=\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{M-1}=\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{10-1}=2.8\phantom{\rule{0.2em}{0ex}}\text{cm}.$

Significance Notation that a greater magnification is accomplished by using a lens with a smaller focal length. We thus need to use a lens with radii of curvature that are less than a few centimeters and hold it very close to our eye. This is not very user-friendly. A compound microscope, explored in the following section, tin can overcome this drawback.

Summary

A simple magnifier is a converging lens and produces a magnified virtual image of an object located within the focal length of the lens.
Athwart magnification accounts for magnification of an image created by a magnifier. It is equal to the ratio of the angle subtended by the epitome to that subtended by the object when the object is observed by the unaided centre.
Angular magnification is greater for magnifying lenses with smaller focal lengths.
Uncomplicated magnifiers can produce every bit groovy every bit tenfold ( $10\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$ ) magnification.

Issues

If the epitome formed on the retina subtends an angle of $30\text{°}$ and the object subtends an angle of $5\text{°}$ , what is the magnification of the image?

$M=6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$

What is the magnification of a magnifying lens with a focal length of x cm if it is held iii.0 cm from the eye and the object is 12 cm from the eye?

How far should you lot hold a ii.1 cm-focal length magnifying glass from an object to obtain a magnification of $10\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$ ? Presume you identify your eye v.0 cm from the magnifying glass.

$\begin{array}{cc}\hfill M& =\left(\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{L}\right)\left(1+\frac{L-\mathcal{\ell }}{f}\right)\hfill \\ \hfill L-\mathcal{\ell }& ={d}_{\text{o}}\hfill \\ \hfill {d}_{\text{o}}& =13\phantom{\rule{0.2em}{0ex}}\text{cm}\hfill \end{array}$

You hold a 5.0 cm-focal length magnifying glass every bit close as possible to your centre. If you have a normal nigh point, what is the magnification?

Yous view a mountain with a magnifying glass of focal length $f=10\phantom{\rule{0.2em}{0ex}}\text{cm}$ . What is the magnification?

$M=2.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$

Y'all view an object by property a 2.5 cm-focal length magnifying glass 10 cm away from information technology. How far from your eye should you hold the magnifying glass to obtain a magnification of $10\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}?$

A magnifying glass forms an image 10 cm on the opposite side of the lens from the object, which is 10 cm away. What is the magnification of this lens for a person with a normal near point if their heart 12 cm from the object?

$\begin{array}{cc}M\hfill & =-2.1\phantom{\rule{0.2em}{0ex}}×\hfill \end{array}$

An object viewed with the naked center subtends a $2\text{°}$ angle. If you view the object through a $10\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$ magnifying glass, what angle is subtended by the image formed on your retina?

For a normal, relaxed eye, a magnifying glass produces an angular magnification of 4.0. What is the largest magnification possible with this magnifying glass?

$\begin{array}{}\\ \\ M=\frac{25\phantom{\rule{0.2em}{0ex}}\text{cm}}{f}\hfill \\ {M}_{\text{max}}=5\hfill \end{array}$

What range of magnification is possible with a 7.0 cm-focal length converging lens?

A magnifying glass produces an angular magnification of 4.5 when used by a young person with a near point of 18 cm. What is the maximum angular magnification obtained by an older person with a near betoken of 45 cm?

$\begin{array}{cc}\hfill {M}_{\text{max}}^{\text{young}}& =1+\frac{18\phantom{\rule{0.2em}{0ex}}\text{cm}}{f}⇒f=\frac{18\phantom{\rule{0.2em}{0ex}}\text{cm}}{{M}_{\text{max}}^{\text{young}}-1}\hfill \\ \hfill {M}_{\text{max}}^{\text{old}}& =9.8\phantom{\rule{0.2em}{0ex}}×\hfill \end{array}$

Glossary

angular magnification: ratio of the angle subtended by an object observed with a magnifier to that observed by the naked center

simple magnifier (or magnifying glass): converging lens that produces a virtual paradigm of an object that is within the focal length of the lens

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Source: https://opentextbc.ca/universityphysicsv3openstax/chapter/the-simple-magnifier/

When One Uses a Magnifying Glass to Read Fine Print One Uses a

16 The Simple Magnifier

Learning Objectives

Summary

Issues

Glossary

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