(No pun intended! (b) The converging nature of the multiple surfaces that make up the eye result in the projection of a real image on the retina.Several important distances appear in Figure 7. [latex]\displaystyle\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\[/latex][latex]\displaystyle\frac{h_{\text{i}}}{h_\text{o}}=\frac{d_{\text{i}}}{d_{\text{o}}}=m\\[/latex]A clear glass light bulb is placed 0.750 m from a convex lens having a 0.500 m focal length, as shown in Figure 9. In this case the virtual image is upright and shrunken. Real images are always inverted, but they can be either larger or smaller than the object. Real images can be projected. Use ray tracing to get an approximate location for the image. So here, there will actually be no image when the object is actually at the focal point. To use a convex lens as a magnifier, the object must be closer to the converging lens than its focal length. Thus the image distance [latex]\frac{1}{d_\text{o}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\[/latex].
The thin lens equations give the most precise results, being limited only by the accuracy of the given information. The magnification is also greater than 1, meaning that the image is larger than the object—in this case, by a factor of 3.

(a) −1.35 m (on the object side of the lens); (b) +10.0; (c) 5.00 cm14. This technique is used in lighthouses and sometimes in traffic lights to produce a directional beam of light from a source that emits light in all directions.Figure 6. A light bulb placed 0.750 m from a lens having a 0.500 m focal length produces a real image on a poster board as discussed in the example above. This is true for movie projectors, cameras, and the eye. [latex]\frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_\text{o}}\\[/latex].Entering known quantities gives a value for[latex]\frac{1}{d_{\text{i}}}\\[/latex]:[latex]\frac{1}{d_{\text{i}}}=\frac{1}{0.500\text{ m}}-\frac{1}{0.750\text{ m}}=\frac{0.667}{\text{m}}\\[/latex]. Virtual image formation has a requirement that the object must be at a distance less than the focal length of the lens. The same formula for the image and object distances used above applies again here. Rays of light entering a converging lens parallel to its axis converge at its focal point F. (Ray 2 lies on the axis of the lens.) The best answers are voted up and rise to the top For a real image, rays from a single source point converge to a single point on the other side of the lens.This means that a point on the image remains well-defined after the optical transformation (refraction by lens).. For a virtual image, rays from a single source point diverge after they pass through the lens. In the world of optics real and virtual images are often described as opposites. The differences between Real and Virtual Images . A virtual image is always upright.
You can use following app to visualize real vs imegenary image.

We can see and photograph virtual images only by using an additional lens to form a real image.Find several lenses and determine whether they are converging or diverging. The two rays continue to diverge on the other side of the lens, but both appear to come from a common point, locating the upright, magnified, virtual image. Ray 1 enters parallel to the axis and exits through the focal point on the opposite side, while ray 2 passes through the center of the lens without changing path. The rules for ray tracing for thin lenses are based on the illustrations already discussed:In some circumstances, a lens forms an obvious image, such as when a movie projector casts an image onto a screen. In order to see an image on a screen light rays must come to a focus on the screen. (a) When a converging lens is held farther away from the face than the lens’s focal length, an inverted image is formed. (The image is virtual.) Sunlight focused by a converging magnifying glass can burn paper. [latex]\frac{1}{d_\text{i}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\[/latex].Entering known values, we obtain a value for [latex]\frac{1}{d_\text{i}}\\[/latex]:[latex]\frac{1}{d_\text{i}}=\frac{1}{10.0\text{ cm}}-\frac{1}{7.50\text{ cm}}=\frac{-0.0333}{\text{cm}}\\[/latex]. Virtual images are always upright and cannot be projected. Physics Stack Exchange works best with JavaScript enabled In this section, we will use the law of refraction to explore the properties of lenses and how they form images.Figure 1. The virtual image is the image that appears at the point where light coming from the object seems to converge after refraction or reflection.


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