Question

To study a tissue sample better, a pathologist holds a \(5.00-\mathrm{cm}\) focal length magnifying glass \(3.00 \mathrm{~cm}\) from the sample. How much magnification can he get from the lens?

Short answer

Answer: The pathologist can achieve a maximum magnification of 2.50 times using the magnifying glass.

Step-by-step solution

Write down the given information

We are given the focal length of the lens, \(f = 5.00 \ \mathrm{cm}\), and the object distance, \(u = 3.00 \ \mathrm{cm}\).

Use the lens formula

The lens formula is given by: \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\), where \(f\) is the focal length, \(u\) is the object distance, and \(v\) is the image distance. We will use this formula to find the image distance.

Find the image distance, \(v\)

Rearrange the lens formula to find \(v\): \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\). Plug in the given values and solve for \(v\): \(\frac{1}{v} = \frac{1}{5.00} - \frac{1}{3.00} = \frac{-2}{15}\) \(v = -7.50\ \mathrm{cm}\) It's normal for \(v\) to be negative in this problem, as it indicates a virtual image being formed on the same side as the object.

Use the magnification formula

The magnification formula is given by: \(M = \frac{-v}{u}\). Use the values of \(u\) and \(v\) we found earlier to calculate the magnification: \(M = \frac{-(-7.50)}{3.00} = 2.50\) The pathologist can get a magnification of 2.50 times from the lens.

Key concepts

Lens Formula

Understanding the lens formula is crucial for studying magnification in optics. This formula, represented as \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \), relates the focal length (f) of the lens to the distance from the object to the lens (u, object distance) and the distance from the lens to the image (v, image distance).

For example, if a lens has a focal length of 5.00 cm and an object is placed 3.00 cm from it, as in our exercise, we can calculate the image distance. By manipulating the formula, solving for v becomes straightforward. This is the essence of Step 2 in the exercise solution, where the calculation provided the image distance necessary to find out the magnification level.

In educational settings, an important tip for grasping this concept is to practice drawing ray diagrams alongside the calculations. Ray diagrams visually demonstrate how light passes through the lens and forms an image, offering a clearer understanding of the relationship between f, u, and v.

Virtual Image

A virtual image is a type of image formed by diverging lenses or through the specific position of objects in converging lenses. Unlike real images, virtual images cannot be projected onto a screen as they appear to be located 'behind' the lens.

The crucial characteristic of virtual images is their orientation and relative position to the lens. In optics, a negative image distance, which we encountered in Step 3 of the solution with a value of v=-7.50 cm, signifies the formation of a virtual image. This means the image will be upright, and located on the same side of the lens as the object.

When studying or teaching optics, highlighting the significance of 'sign conventions' is vital for avoiding confusion. In diagrams, virtual images are often represented with dashed lines, illustrating that they are not formed by the actual convergence of light rays.

Optics in Physics

Optics, the segment of physics concerned with the behavior and properties of light (including its interactions with matter and the construction of instruments that use or detect it), is a broad field encompassing various concepts such as reflection, refraction, magnification, and more.

Our example problem, involving a pathologist using a magnifying glass to better examine a tissue sample, touches upon the practical application of lens-based magnification. Within this framework, optics is used to describe how lenses can magnify objects, a fundamental aspect of microscopes, telescopes, eyeglasses, cameras, etc.

For educational content, focusing on real-world applications helps to illustrate theoretical principles in a tangible way. Providing examples from everyday life can enhance a student’s comprehension and interest in the subject of physics. Experimentation with actual lenses and observing the formation of real and virtual images is also highly recommended for an intuitive understanding of optics.