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The magnification of an image is both the hi / ho ratio and the -di / do ratio. Setting the -di / do ratio equal to -2 allows one to determine the image distance:

The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:

10. The focal point is located 20.0 cm from a double concave lens. An object is placed 12 cm from the lens. Determine the image distance.

The negative value for image distance indicates that the image is a virtual image located on the object's side of the lens. Again, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image distance, a negative value always means the image is located on the object's side of the lens. Note also that the image height is a positive value, meaning an upright image. Any image that is upright and located on the object's side of the lens is considered to be a virtual image.

5. A magnified, inverted image is located a distance of 32.0 cm from a double convex lens with a focal length of 12.0 cm. Determine the object distance and tell whether the image is real or virtual.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

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To determine the image distance, the lens equation will have to be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

From the calculations in the second sample problem it can be concluded that if a 4.00-cm tall object is placed 8.30 cm from a double convex lens having a focal length of 15.2 cm, then the image will be enlarged, upright, 8.81-cm tall and located 18.3 cm from the lens on the object's side. The results of this calculation agree with the principles discussed earlier in this lesson. In this case, the object is located in front of the focal point (i.e., the object distance is less than the focal length) and the image is located behind the lens. This falls into the category of Case 5: The object is located in front of F (for a converging lens).

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Like many mathematical problems in physics, the skill is only acquired through much personal practice. Perhaps you would like to take some time to try the following problems.

From 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. The results of this calculation agree with the principles discussed earlier in this lesson. 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. This falls into the category of Case 1: The object is located beyond 2F for a converging lens.

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2. Determine the image distance and image height for a 5-cm tall object placed 30.0 cm from a double convex lens having a focal length of 15.0 cm.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

From the calculations in this problem it can be concluded that if a 4.00-cm tall object is placed 35.5 cm from a diverging lens having a focal length of 12.2 cm, then the image will be upright, 1.02-cm tall and located 9.08 cm from the lens on the object's side. The results of this calculation agree with the principles discussed earlier in this lesson. Diverging lenses always produce images that are upright, virtual, reduced in size, and located on the object's side of the lens.

8. Determine the focal length of a double concave lens that produces an image that is 16.0 cm behind the lens when the object is 28.5 cm from the lens.

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1. Determine the image distance and image height for a 5-cm tall object placed 45.0 cm from a double convex lens having a focal length of 15.0 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.

4. Determine the image distance and image height for a 5-cm tall object placed 10.0 cm from a double convex lens having a focal length of 15.0 cm.

To determine the image distance, the lens equation must be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

3. Determine the image distance and image height for a 5-cm tall object placed 20.0 cm from a double convex lens having a focal length of 15.0 cm.

The negative values for image distance indicate that the image is located on the object's side of the lens. As mentioned, a negative or positive sign in front of the numerical value for a physical quantity represents information about direction. In the case of the image distance, a negative value always indicates the existence of a virtual image located on the object's side of the lens. In the case of the image height, a positive value indicates an upright image.

The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

9. A 2.8-cm diameter coin is placed a distance of 25.0 cm from a double concave lens that has a focal length of -12.0 cm. Determine the image distance and the diameter of the image.

To determine the image distance, the lens equation will have to be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

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The numerical values in the solution above were rounded when written down, yet unrounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

7. A double concave lens has a focal length of -10.8 cm. An object is placed 32.7 cm from the lens's surface. Determine the image distance.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

The negative values for image height indicate that the image is an inverted image. 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. In the case of the image height, a negative value always indicates an inverted image.

6. ZINGER: An inverted image is magnified by 2 when the object is placed 22 cm in front of a double convex lens. Determine the image distance and the focal length of the lens.

As a demonstration of the effectiveness of the lens equation and magnification equation, consider the following sample problem and its solution.