Understanding Magnification in Real Inverted Images Formed by Concave Mirrors

When discussing the properties of lenses and mirrors, one common question involves the magnification of real images in concave mirrors. It is often presented as a challenge whether a real inverted image can have a positive magnification. In this article, we delve into this phenomenon and explore the circumstances under which this can occur.

Magnification in Real Inverted Images by Concave Mirrors

A real inverted image can indeed have a positive magnification when the object is placed between the focal point and the mirror. This configuration leads to an inverted, real image that is magnified. The magnification formula for a concave mirror is given by:

m -v/u

Where:

v is the distance of the image from the mirror. u is the distance of the object from the mirror.

Let's break down the scenario: when the object is placed between the focal point and the mirror, v (image distance) is positive, and u (object distance) is negative. This results in a positive magnification, indicating that the image is larger than the object. However, it is important to note that the image is still inverted and real.

Understanding the Sign Convention

One critical point to remember is that the magnification of a real image (formed by a concave mirror) is always negative. This is due to the object and image being on opposite sides of the lens or mirror. For a virtual image, however, the sign of the magnification is positive.

Common Misconception

Sometimes, students might encounter questions where the teacher assumes that the magnification should be calculated based on the real image scenario, even if the question pertains to virtual images. For instance, if a student is asked to calculate magnification for a real image, they should consider the formula m -v/u to confirm the negative magnification.

Example Problem

A common problem involves calculating the distance of an object from a concave mirror to form a real image of magnification 2. Given that the mirror's radius is 20 cm, we can solve this problem as follows:

1. Since it is a mirror, the radius of curvature R is 20 cm. Using the formula R 2f, we find the focal length f is -10 cm (negative because the focal point is on the same side as the object).

2. The magnification formula M -v/u, with M -2 for a real image, means v 2u.

3. Substituting v 2u into the lens formula 1/f 1/v 1/u, we can solve for u.

4. After calculations, we get u -15 cm, and v -30 cm. The negative sign confirms that the image is on the same side as the object and not behind the mirror.

5. The magnification M -2, indicating that the image height hi is twice the object height ho and inverted.

By following these steps, we can accurately determine the magnification and image location for real inverted images formed by concave mirrors.

Conclusion

In summary, a real inverted image can indeed have a positive magnification in specific circumstances, such as when the object is placed between the focal point and the mirror. Understanding the sign convention and applying the appropriate formulas are crucial for solving related problems in optics. Always consider the correct sign for magnification when dealing with real images formed by concave mirrors.