Question

An object of height \(h\) is placed at a distance \(d_{0}\) on the left side of a converging lens of focal length \(f\left(f

Short answer

Question: An object is placed in front of a converging lens, and an image is formed 3 times the focal length, f, to the right of the lens. a) Calculate the distance the object needs to be placed on the left side of the lens, d₀. b) Calculate the magnification, M, of the image. Answer: a) The object must be placed at a distance of \(\frac{3f}{2}\) on the left side of the lens. b) The magnification of the image is -2, meaning the image is flipped and twice the size of the object.

Step-by-step solution

Solving for \(d_0\) with the lens equation

We have the lens equation: \(\frac{1}{f} = \frac{1}{d_0} + \frac{1}{d}\). Given that \(d = 3f\), we can substitute this into the lens equation and solve for \(d_0\): \(\frac{1}{f} = \frac{1}{d_0} + \frac{1}{3f}\) Now, we'll solve for \(d_0\): \(\frac{1}{d_0} =\frac{1}{f} - \frac{1}{3f} = \frac{3-1}{3f} = \frac{2}{3f}\) Therefore, \(d_0 = \frac{3f}{2}\) So the object must be placed at a distance \(\frac{3f}{2}\) on the left side of the lens for the image to form at a distance \(3f\) on the right side of the lens.

Calculating the magnification

Now that we have found \(d_0 = \frac{3f}{2}\), we can use the formula for magnification, \(M = \frac{-d}{d_0}\): \(M = \frac{-3f}{\frac{3f}{2}}\) Multiplying by \(\frac{2}{1}\) to clear the fraction: \(M = \frac{-2(3f)}{1(3f)}\) Canceling the common factor of \(3f\) in the numerator and denominator: \(M = -2\) The magnification is -2, indicating that the image is flipped (which is denoted by the negative sign) and is twice the size of the object.

Key concepts

Lens Equation

The lens equation is a fundamental formula used in optics, helping us understand how converging lenses, like the one in this exercise, form images. It is represented by the equation \( \frac{1}{f} = \frac{1}{d_0} + \frac{1}{d} \), where:

• \( f \) is the focal length of the lens,
• \( d_0 \) is the distance from the lens to the object, and
• \( d \) is the distance from the lens to the image.

Knowing how to rearrange and use this equation can help solve many problems related to lenses. For instance, if the desired image distance \( d \) and the lens's focal length \( f \) are known, you can rearrange the equation to find the object distance \( d_0 \). In our case, we needed the image to form at \( 3f \). Substituting this into the lens equation allowed us to solve for \( d_0 \), leading to \( d_0 = \frac{3f}{2} \). This indicates how far the object should be placed from the lens to achieve the required image distance.

Magnification

Magnification tells us how much larger or smaller the image is compared to the object. It also gives us information about the orientation of the image. In optics, magnification \( M \) can be calculated using the formula: \( M = \frac{-d}{d_0} \), where:

• \( d \) is the image distance,
• \( d_0 \) is the object distance, and
• The negative sign indicates that a negative magnification means the image is inverted.

In our exercise, with \( d = 3f \) and \( d_0 = \frac{3f}{2} \), plugging these into the magnification formula gives \( M = -2 \). This result tells us two things:

• The image is flipped upside down as the magnification is negative, and
• It is twice the size of the original object.

Image Formation

Image formation through lenses is a key concept in understanding how lenses function in various applications, from eyeglasses to cameras. A converging lens, as used in our problem, bends light rays inward, converging them at a point to form an image.
Images formed by lenses can be either real or virtual. A real image is formed when the light actually converges at the image location and can be projected onto a screen. Conversely, a virtual image forms when the light rays do not actually meet but only appear to converge.
In the given exercise, the image formed at a distance \( 3f \) on the right side of the lens is real and inverted, which is confirmed by the negative magnification value. Understanding how the arrangement of the object, lens, and placement affects the type and orientation of the image is crucial for effectively utilizing lenses in practical scenarios.