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The figure shows a device for constructing a realistic optical illusion. Two mirrors of equal focal length are put against each other with their silvered surfaces facing inward. A small object placed in the bottom of the cavity will have its image projected in the air above. The way it works is that the top mirror produces a virtual image, and the bottom mirror then creates a real image of the virtual image. (a) Show that if the image is to be positioned as shown, at the mouth of the cavity, then the focal length of the mirrors is related to the dimension \(h\) via the equation $$\frac{1}{f}=\frac{1}{h}+\frac{1}{h+\left(\frac{1}{h}-\frac{1}{f}\right)^{-1}}$$ (b) Restate the equation in terms of a single variable \(x=h / f\), and show that there are two solutions for \(x\). Which solution is physically consistent with the assumptions of the calculation?

Step by step solution

01

Understand the Mirror Equation

For a mirror, the mirror equation is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. In this setup, the small object is at the bottom of the cavity, and we want its image at the mouth of the cavity.

02

Analyze Virtual and Real Images

The top mirror creates a virtual image at a distance \( d_i = 2h \) (as the object is placed at \( h \)), which serves as the object for the bottom mirror. For the bottom mirror, \( d_o = 2h \). The real image should also be at \( 2h \), above the bottom of the cavity.

03

Substitute into the Mirror Equation

Substitute into the mirror equation for the bottom mirror: \( \frac{1}{f} = \frac{1}{2h} + \frac{1}{d_i} \). We need the image, \( d_i \), to appear at \( 2h \). As we analyze the behavior of images, \( \frac{1}{f} = \frac{1}{h} + \frac{1}{h + (\frac{1}{h} - \frac{1}{f})^{-1}} \) describes this setup by considering the alignment needed to place the real image at the mouth.

04

Restate the Equation using \(x = \frac{h}{f}\)

Replace \( h \) with \( xf \), so the given equation becomes and manipulate it: \[\frac{1}{f} = \frac{1}{xf} + \frac{1}{xf + \left(\frac{1}{xf} - \frac{1}{f}\right)^{-1}}\]. Simplifying gives an equation involving \( x \).

05

Solve for x

Rearrange the equation to solve for \( x \), by clearing fractions and simplifying, we find: \[x^2 - 3x + 1 = 0\]. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) with \( a = 1, b = -3, c = 1 \).

06

Apply the Quadratic Formula

The solutions for \( x \) obtained from the quadratic formula are \( x = \frac{3 \pm \sqrt{5}}{2} \). So, the two solutions are \( x = \frac{3 + \sqrt{5}}{2} \) and \( x = \frac{3 - \sqrt{5}}{2} \).

07

Determine the Physically Consistent Solution

Since \( x = \frac{h}{f} \) is a ratio, it must be positive and consistent with the mirror setup. \( x = \frac{3 - \sqrt{5}}{2} \) is less than 1, which does not make sense as it suggests a focal length greater than height, which contradicts the image setup. Therefore, \( x = \frac{3 + \sqrt{5}}{2} \) is the physically consistent solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mirror Equation

The mirror equation is essential for understanding how images form in mirror systems. It is given by: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Here, \( f \) represents the focal length of the mirror, \( d_o \) is the object distance (the distance from the object to the mirror), and \( d_i \) is the image distance (the distance from the image to the mirror).
In the problem under consideration, two mirrors of equal focal length are used to create optical illusions. By using this equation, we can determine the relationships between these distances to predict where the image will appear. The concept helps ensure that images are positioned at desired locations, crucial for constructing effective illusions.

Virtual Image

A virtual image is one that appears to be located at a position from which light does not actually come. When you look into a mirror, the reflection you see is a virtual image. Light rays appear to diverge from the image, but they do not converge in reality.
A key characteristic of the virtual image is that it cannot be projected onto a screen because the light does not physically arrive there. In the described setup, the top mirror creates a virtual image at a distance \( 2h \) from the object. This virtual image acts as the subject for the bottom mirror, which works to convert the virtual into a real image, making the optical illusion possible.

Real Image

Real images are formed when light converges at a point and can be projected onto a screen. Unlike virtual images, real images carry all the actual light from the object, focused into an image point.
In the setup, once the virtual image is created by the top mirror, the bottom mirror takes this image as its object. By applying the principles of the mirror equation, the bottom mirror produces a real image at the mouth of the cavity. This sequence of virtual to real image creation is key in many optical illusion devices where an object appears to magically float in mid-air.

Focal Length

The focal length of a mirror is the distance between the mirror and its focal point, where parallel light rays converge. It is a pivotal property that dictates how a mirror directs light rays and creates images.
In our problem, both mirrors have the same focal length, and they rely on this characteristic to align images properly. The focal length determines how the virtual image formed by the first mirror turns into a real image by the second mirror. By manipulating the mirror equation, we relate focal length to the dimension \( h \), ensuring the image is precisely at the mouth of the cavity. This understanding is crucial for fine-tuning the placement of images in optical devices.