Example: A particular nearsighted person is unable to see objects clearly when they are beyond 2.5 m away (the far point of this particular eye). What should the focal length be in a lens prescribed to correct this problem? Solution: The purpose of the lens in this instance is to move an object from infinity to a distance where it can be seen clearly .This is accomplished by having the lens

produce an image at the far point. From the thin lens equation, we have We use a negative sign for the image distance because the image is virtual and in front of the eye. As you should have suspected, the lens must be a diverging lens (one with a negative focal length) to correct nearsightedness. The Simple Magnifier Suppose an object is viewed at some distance p from the eye, as illustrated in Figure 36.42. The size of the image formed at the

retina depends on the angle subtended by the object at the eye. As the object moves closer to the eye, increases and a larger image is observed. However, an average normal eye cannot focus on an object closer than about 25 cm, the near point (Fig. 36.43a). Therefore, is maximum at the near point. To further increase the apparent angular size of an object, a converging lens can be placed in front of the eye as in Figure 36.43b, with the object located at point O, just inside the focal point of the lens.

We define angular magnification m as the ratio of the angle subtended by an object with a lens in use to the angle subtended by the object placed at the near point with no lens in use. The angular magnification is a maximum when the image is at the near point of the eyethat is, when q=- 25 cm. The object distance corresponding to this image distance can be calculated from the thin lens equation: where f is the focal length of the magnifier in centimeters If we make the small-angle approximations

Then 36.20 becomes Although the eye can focus on an image formed anywhere between the near point and infinity, it is most relaxed when the image is at infinity. For the image formed by the magnifying lens to appear at infinity, the object has to be at the focal point of the lens. In this case, Equations 36.21 become and the magnification is With a single lens, it is possible to obtain angular magnifications up to about 4 without serious aberrations. Magnifications up to

about 20 can be achieved by using one or two additional lenses to correct for aberrations. Example What is the maximum magnification that is possible with a lens having a focal length of 10 cm, and what is the magnification of this lens when the eye is relaxed? Solution The maximum magnification occurs when the image is located at the near point of the eye. Under these circumstances, Equation 36.22 gives When the eye is relaxed, the image is at infinity. In this case, we use Equation 36.23:

Example A lens that has a focal length of 5.00 cm is used as a magnifying glass. (a) To obtain maximum magnification, where should the object be placed? (b) What is the magnification? Solution: The Compound Microscope A simple magnifier provides only limited assistance in inspecting

minute details of an object. Greater magnification can be achieved by combining two lenses in a device called a compound microscope It consists of one lens, the objective, that has a very short focal length fo<1 cm and a second lens, the eyepiece, that has a focal length fe of a few centimeters. The two lenses are separated by a distance L that is much greater than either fo or fe . The object, which is placed just outside the focal point of the objective, forms a real, inverted image at I1, and this

image is located at or close to the focal point of the eyepiece. The eyepiece, which serves as a simple magnifier, produces at I2 a virtual, enlarged image of I1. The lateral magnification M1 of the first image is -q1/p1. that q1 is approximately equal to L and that the object is very close to the focal point of the objective: p1 fo. Thus, the lateral magnification by the objective is The angular magnification by the eyepiece for an object (corresponding to the image at I1) placed at the focal point of the eyepiece is, from Equation 36.23,

The overall magnification of the image formed by a compound microscope is defined as the product of the lateral and angular magnifications: The negative sign indicates that the image is inverted. The microscope has extended human vision to the point where we can view previously unknown details of incredibly small objects. Example:

The distance between eyepiece and objective lens in a certain compound microscope is 23.0 cm. The focal length of the eyepiece is 2.50 cm, and that of the objective is 0.400 cm. What is the overall magnification of the microscope? Solution: Example: The desired overall magnification of a compound microscope is 140x. The objective alone produces a lateral magnification of

12.0x. Determine the required focal length of the eyepiece. Solution: The Telescope Two fundamentally different types of telescopes exist. The refracting telescope uses a combination of lenses to form an image, this telescope has an objective and an eyepiece. The two lenses are arranged so that the objective forms a real,

inverted image of a distant object very near the focal point of the eyepiece. Because the object is essentially at infinity, this point at which I 1 forms is the focal point of the objective. The eyepiece then forms, at I 2, an enlarged, inverted image of the image at I 1 In order to provide the largest

possible magnification, the image distance for the eyepiece is infinite. This means that the light rays exit the eyepiece lens parallel to the principal axis, and the image of the objective lens must form at the focal point of the eyepiece. Hence, the two lenses are separated by a distance fo+fe , which corresponds to the length of the

telescope tube. The angular magnification of the telescope is given by /o , where o is the angle subtended by the object at the objective and is the angle subtended by the final image at the viewers eye. The angle o (to the left of the objective subtended by the first image at the objective. Thus, where the negative sign indicates that the image is inverted.

The angle subtended by the final image at the eye is the same as the angle that a ray coming from the tip of I1 and traveling parallel to the principal axis makes with the principal axis after it passes through the lens. Thus I1, and both it and I2 point in the same direction. the angular magnification of the telescope can be expressed as the angular magnification of a telescope equals the ratio of the objective focal length to the eyepiece focal length. The negative sign indicates that the image is inverted.

When we look through a telescope at such relatively nearby objects as the Moon and the planets, magnification is important. However, individual stars in our galaxy are so far away that they always appear as small points of light no matter how great the magnification A large research telescope that is used to study very distant objects must have a great diameter to gather as much light as possible. The difficulties for building telescopes: It is expensive to manufacture large lenses for refracting telescopes.

Another difficulty with large lenses is that their weight leads to sagging, which is an additional source of aberration.