Question

A classmate claims that by using a \(40.0-\mathrm{cm}\) focal length mirror, he can project onto a screen a \(10.0-\mathrm{cm}\) tall bird locat ed 100 . \(\mathrm{m}\) away. He claims that the image will be no less than \(1.00 \mathrm{~cm}\) tall and inverted. Will he make good on his claim?

Short answer

Answer: No, the student's claim is partially correct. Though the projected image is indeed inverted, it is smaller than 1 cm tall (approximately 0.421 cm).

Step-by-step solution

Identify the given information

We are given: 1. Focal length of the mirror (f) = 40 cm 2. Object distance from the mirror (d_o) = 100 m (convert it to cm) 3. Object height (h_o) = 10 cm We need to find the height of the image (h_i) and verify if it is inverted.

Convert the distance to a matching unit

Convert the object distance from meters to centimeters: d_o = 100 m * 100 cm/m = 10,000 cm

Use the mirror equation to find the image distance (d_i)

The mirror equation is: 1/f = 1/d_o + 1/d_i Rearrange to solve for d_i: 1/d_i = 1/f - 1/d_o Substitute the values: 1/d_i = 1/(40 cm) - 1/(10,000 cm) Now, calculate the value of d_i: d_i = 1/((1/40 - 1/10000)) ≈ 42.11 cm

Use the magnification equation to find the image height (h_i)

The magnification equation is: M = h_i / h_o = -d_i / d_o We want to find h_i, so rearrange the equation: h_i = h_o * (-d_i / d_o) Substitute the values: h_i = (10 cm) * (-42.11 cm / 10,000 cm) Calculate the value of h_i: h_i ≈ -0.421 cm

Determine if the claim is correct

From our calculations, we find that the height of the image (h_i) is approximately -0.421 cm, which means the image is indeed inverted (negative sign). However, the image is smaller than the claimed 1 cm (0.421 cm < 1 cm). Therefore, the claim is partially correct - the image is inverted, but it is smaller than 1 cm.

Key concepts

Mirror Equation

The mirror equation is an essential tool in geometrical optics, particularly useful when dealing with curved mirrors. It establishes the relationship between the focal length of the mirror, the object distance, and the image distance. The equation is given by:

• \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)

Here, \( f \) is the focal length of the mirror, \( d_o \) is the distance from the object to the mirror, and \( d_i \) is the distance from the image to the mirror. This equation allows us to locate the position of the image formed by the mirror.

In the given exercise, we have focal length \( f = 40 \) cm and object distance \( d_o = 10,000 \) cm. By substituting these values into the mirror equation and solving for \( d_i \), we find the image distance. In this example, the calculation yields an image distance of approximately \( d_i = 42.11 \) cm.

This equation is a fundamental part of understanding how mirrors project images, whether onto a screen or another surface.

Magnification Equation

The magnification of an image in optics tells us how much larger or smaller the image is compared to the object itself. This is described by the magnification equation, which relates the image height, object height, image distance, and object distance:

• \( M = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \)

Here, \( M \) is the magnification, \( h_i \) is the image height, \( h_o \) is the object height, \( d_i \) is the image distance, and \( d_o \) is the object distance. The negative sign indicates the inverted nature of the image for real images formed by concave mirrors.

Using the values found in the exercise, we calculate the image height \( h_i \) as approximately -0.421 cm. This reflects both the image's size and its orientation - it's shorter and inverted compared to the original object.

This equation helps us predict not just the size but also whether an image will appear inverted or upright, crucial for various practical applications.

Image Formation

Image formation by mirrors involves using light reflection principles. When light rays from an object reflect off a mirror, they may converge to form an image. There are two primary scenarios: real images and virtual images.

Real images are formed on the same side as the object when the light rays actually converge after reflection. These are the types usually projected onto a screen, like in the given exercise, and appear inverted.

In the exercise, with a focal length of 40 cm, the mirror reflects rays from the bird onto a screen, forming a real and inverted image approximately 42.11 cm from the mirror. This occurs because the focal length and distances satisfy the conditions given by the mirror equation.

Understanding how images form isn't just theoretical—it's practical and used in everyday mirrors and optical devices like cameras and telescopes.

Inverted Image

One of the intriguing aspects of geometrical optics is the formation of inverted images. In mirrors, an inverted image occurs when the top and bottom of the object are flipped in the image.

In the exercise, the calculated negative sign for the image height \( h_i \) indicates the image is inverted. In the language of optics, this sign convention tells us the image's vertical orientation. For concave mirrors, real images formed beyond the focal point are always inverted, meaning they have a top-to-bottom reversal.

This inverted nature is especially useful for certain applications like microscopes and cameras, where the orientation might need correction in processing. It's also why the statement in the exercise holds partly true; while the image is inverted, it falls short of the claimed height. Thus, understanding whether an image is inverted or upright is critical in evaluating optical setups.