Now lets try a second sample problem: Sample Problem #2 A 4.00-cm tall light bulb is placed a distance of 8.30 cm from a double convex lens having a focal length of 15.2 cm. This falls into the category of Case 1: The object is located beyond 2F for a converging lens. In this case, the object is located beyond the 2F point (which would be two focal lengths from the lens) and the image is located between the 2F point and the focal point. The results of this calculation agree with the principles discussed earlier in this lesson. In the case of the image height, a negative value always indicates an inverted image.įrom the calculations in this problem it can be concluded that if a 4.00-cm tall object is placed 45.7 cm from a double convex lens having a focal length of 15.2 cm, then the image will be inverted, 1.99-cm tall and located 22.8 cm from the lens. As is often the case in physics, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction.
The negative values for image height indicate that the image is an inverted image. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. To determine the image height, the magnification equation is needed. The final answer is rounded to the third significant digit. The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The following lines represent the solution to the image distance substitutions and algebraic steps are shown. To determine the image distance, the lens equation must be used. Next identify the unknown quantities that you wish to solve for. Like all problems in physics, begin by the identification of the known information. Determine the image distance and the image size. Sample Problem #1 A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.Īs a demonstration of the effectiveness of the lens equation and magnification equation, consider the following sample problem and its solution.
The magnification equation is stated as follows: The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (h i) and object height (h o). The lens equation expresses the quantitative relationship between the object distance (d o), the image distance (d i), and the focal length (f). To obtain this type of numerical information, it is necessary to use the Lens Equation and the Magnification Equation.
While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and image size. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. The use of these diagrams was demonstrated earlier in Lesson 5 for both converging and diverging lenses. Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a lens.
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