Phys102 Lecture 23/24 Lenses and Optical Instruments Key Points Thin Lenses; Ray Tracing Combinations of Lenses The Human Eye; Corrective Lenses Compound Microscope References 23-7,8,9; 25-2,3,4,5.

Thin Lenses Thin lenses are those whose thickness is small compared to their radius of curvature. They may be either converging (a) or diverging (b). Figure 33-2. (a) Converging lenses and (b) diverging lenses, shown in cross section. Converging lenses are thicker in the center whereas diverging lenses are thinner in the center. (c) Photo of a converging lens (on the left) and a diverging lens (right). (d) Converging lenses (above), and diverging lenses (below), lying flat, and raised off the paper to form images.

Ray Tracing Parallel rays are brought to a focus by a converging lens (one that is thicker in the center than it is at the edge). Figure 33-3. Parallel rays are brought to a focus by a converging thin lens.

Ray Tracing A diverging lens (thicker at the edge than in the center) makes parallel light diverge; the focal point is that point where the diverging rays would converge if projected back. Figure 33-5. Diverging lens.

The Power of a Lens The power of a lens is the inverse of its focal length: Lens power is measured in diopters, D: 1 D = 1 m-1.

Thin Lenses; Ray Tracing Ray tracing for thin lenses is similar to that for mirrors. We have three key rays: This ray comes in parallel to the axis and exits through the focal point. This ray comes in through the focal point and exits parallel to the axis. This ray goes through the center of the lens and is undeflected.

Figure 33-6. Finding the image by ray tracing for a converging lens Figure 33-6. Finding the image by ray tracing for a converging lens. Rays are shown leaving one point on the object (an arrow). Shown are the three most useful rays, leaving the tip of the object, for determining where the image of that point is formed.

For a diverging lens, we can use the same three rays. Figure 33-8. Finding the image by ray tracing for a diverging lens. Ray 1 is drawn parallel to the axis, but does not pass through the focal point F’ behind the lens. Instead it seems to come from the focal point F in front of the lens.

Conceptual Example: Half-blocked lens. What happens to the image of an object if the top half of a lens is covered by a piece of cardboard? Half of the image is blocked, so we can only see half of the image. We can still see the whole image, but not as bright. The image is not affected at all. Solution: The image is unchanged (follow the rays); only its brightness is diminished, as some of the light is blocked.

The Thin Lens Equation The thin lens equation is similar to the mirror equation: Figure 33-9. Deriving the lens equation for a converging lens.

The sign conventions are slightly different: The focal length is positive for converging lenses and negative for diverging. The object distance is positive when the object is on the same side as the light entering the lens (not an issue except in compound systems); otherwise it is negative. The image distance is positive if the image is on the opposite side from the light entering the lens; otherwise it is negative. The height of the image is positive if the image is upright and negative otherwise.

The magnification formula is also the same as that for a mirror: The power of a lens (p=1/f) is positive if it is converging and negative if it is diverging.

Problem Solving: Thin Lenses Draw a ray diagram. The image is located where the key rays intersect. Solve for unknowns. Follow the sign conventions. Check that your answers are consistent with the ray diagram.

Example: Image formed by converging lens. What are (a) the position, and (b) the size, of the image of a 7.6-cm-high leaf placed 1.00 m from a +50.0-mm-focal-length camera lens? Solution: The figure shows the appropriate ray diagram. The thin lens equation gives di = 5.26 cm; the magnification equation gives the size of the image to be -0.40 cm. The signs tell us that the image is behind the lens and inverted.

Example: Object close to converging lens. An object is placed 10 cm from a 15-cm-focal-length converging lens. Determine the image position and size (a) analytically, and (b) using a ray diagram. Figure 33-12. An object placed within the focal point of a converging lens produces a virtual image. Example 33–3. Solution: a. The thin lens equation gives di = -30 cm and m = 3.0. The image is virtual, enlarged, and upright. b. See the figure.

Example: Diverging lens. Where must a small insect be placed if a 25-cm-focal-length diverging lens is to form a virtual image 20 cm from the lens, on the same side as the object? Solution: Since the lens is diverging, the focal length is negative. The lens equation gives do = 100 cm.

Combinations of Lenses In lens combinations, the image formed by the first lens becomes the object for the second lens (this is where object distances may be negative). The total magnification is the product of the magnification of each lens.

Example: A two-lens system. Two converging lenses, A and B, with focal lengths fA = 20.0 cm and fB = 25.0 cm, are placed 80.0 cm apart. An object is placed 60.0 cm in front of the first lens. Determine (a) the position, and (b) the magnification, of the final image formed by the combination of the two lenses. Figure 33-14. Two lenses, A and B, used in combination. The small numbers refer to the easily drawn rays. Solution: a. Using the lens equation we find the image for the first lens to be 30.0 cm in back of that lens. This becomes the object for the second lens - it is a real object located 50.0 cm away. Using the lens equation again we find the final image is 50 cm behind the second lens. b. The magnification is the product of the magnifications of the two lenses: +0.500. The image is half the size of the object and upright.

Example: Measuring f for a diverging lens. To measure the focal length of a diverging lens, a converging lens is placed in contact with it. The Sun’s rays are focused by this combination at a point 28.5 cm behind the lenses as shown. If the converging lens has a focal length fC of 16.0 cm, what is the focal length fD of the diverging lens? Assume both lenses are thin and the space between them is negligible. Figure 33-15. Determining the focal length of a diverging lens. Example 33–6. Solution: The image distance for the first lens is its focal length (the object is essentially infinitely far away). This forms the (virtual) object for the second lens. Using the lens equation and the position of the focused rays, and solving for the focal length of the second lens, gives -36.5 cm.

Cameras Basic parts of a camera: Lens Light-tight box Shutter Film or electronic sensor Figure 33-18. A simple camera.

The Human Eye The human eye resembles a camera in its basic functioning, with an adjustable lens, the iris, and the retina. Figure 33-25. Diagram of a human eye.

The Human Eye Near point: closest distance at which eye can focus clearly. Normal is about 25 cm. Far point: farthest distance at which object can be seen clearly. Normal is at infinity. Nearsightedness: far point is too close. Farsightedness: near point is too far away.

Corrective Lenses Nearsightedness can be corrected with a diverging lens. Figure 33-27a. Correcting eye defects with lenses: (a) a nearsighted eye, which cannot focus clearly on distant objects, can be corrected by use of a diverging lens

Corrective Lenses And farsightedness with a converging lens. Figure 33-27b. (b) a farsighted eye, which cannot focus clearly on nearby objects, can be corrected by use of a converging lens.

Example: Farsighted eye. Sue is farsighted with a near point of 100 cm. Reading glasses must have what lens power so that she can read a newspaper at a distance of 25 cm? Assume the lens is very close to the eye. Solution: Using do = 25 cm and di = -100 cm gives f = 0.33 m. The lens power is +3.0 D.

Example: Nearsighted eye. A nearsighted eye has near and far points of 12 cm and 17 cm, respectively. (a) What lens power is needed for this person to see distant objects clearly, and (b) what then will be the near point? Assume that the lens is 2.0 cm from the eye (typical for eyeglasses). Solution: The lens should put a distant object at the far point of the eye. Using do = ∞ and di = -15 cm (since the eye is 2 cm from the lens) gives f = -0.15 m. The lens power is -6.7 D.

Magnifying Glass A magnifying glass (simple magnifier) is a converging lens. It allows us to focus on objects closer than the near point, so that they make a larger, and therefore clearer, image on the retina. Figure 33-33a. Leaf viewed (a) through a magnifying glass. The eye is focused at its near point

The power of a magnifying glass is described by its angular magnification: If the eye is relaxed (N is the near point distance and f the focal length): If the eye is focused at the near point:

Example: A jeweler’s “loupe.” An 8-cm-focal-length converging lens is used as a “jeweler’s loupe,” which is a magnifying glass. Estimate (a) the magnification when the eye is relaxed, and (b) the magnification if the eye is focused at its near point N = 25 cm. Solution: a. With the relaxed eye focused at infinity, M = N/f, which is about a factor of 3. b. With the eye focused at its near point, M = 1 + N/f, or about a factor of 4.

Compound Microscope A compound microscope also has an objective and an eyepiece; it is different from a telescope in that the object is placed very close to the eyepiece. Figure 33-40. Compound microscope: (a) ray diagram, (b) photograph (illumination comes from the lower right, then up through the slide holding the object).

Compound Microscope The magnification is given by

Example: Microscope. A compound microscope consists of a 10X eyepiece and a 50X objective 17.0 cm apart. Determine (a) the overall magnification, (b) the focal length of each lens, and (c) the position of the object when the final image is in focus with the eye relaxed. Assume a normal eye, so N = 25 cm. Solution: a. The overall magnification is 500X. b. The eyepiece focal length is N/Me = 2.5 cm. Then, solving equation 33-8 for do gives do = 0.29 cm. Finally, the thin lens equation gives fo = 0.28 cm. c. See (b). do = 0.29 cm.