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Chapter 5: Problem 86

A homemade telephoto "lens" (Fig. P.5.86) consists of two spherical mirrors. The radius of curvature is \(2.0 \mathrm{m}\) for the primary (the big mirror) and \(60 \mathrm{cm}\) for the secondary (the small mirror). How far from the smaller mirror should the film plane be located if the object is a star? What is the effective focal length of the system?

Short Answer

Expert verified
Film plane distance is 0.871 m, and system's focal length is 0.429 m.

Step by step solution

01

Understand the Mirror System

The telephoto lens system consists of two mirrors: a large primary mirror with a radius of curvature of \( R_1 = 2.0 \) m and a smaller secondary mirror with a radius of curvature of \( R_2 = 0.6 \) m. The focal length \( f \) of a mirror is given by \( f = \frac{R}{2} \).
02

Calculate Focal Lengths of Individual Mirrors

For the primary mirror, calculate its focal length: \[ f_1 = \frac{R_1}{2} = \frac{2.0}{2} = 1.0 \text{ m} \]. For the secondary mirror, calculate its focal length: \[ f_2 = \frac{R_2}{2} = \frac{0.6}{2} = 0.3 \text{ m} \].
03

Calculate the Image Position of the Primary Mirror

Since the object is a star located very far away, the object distance for the primary mirror is essentially infinity. According to the mirror equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o = \infty \), we have \( d_i = f_1 = 1.0 \text{ m} \). Thus, the image of the star is formed 1.0 m away from the primary mirror.
04

Determine the Effective Focal Length of the System

The image formed by the primary mirror acts as the object for the secondary mirror. The separation \( d \) between the mirrors is the sum of their focal lengths, \( d = f_1 + f_2 = 1.0 + 0.3 = 1.3 \text{ m} \). The effective focal length \( f \) of the system is determined by the lens formula for a two-lens/mirror system: \( \frac{1}{f} = \frac{1}{f_1} - \frac{1}{f_2} \), resulting in \[ f = \frac{f_1 f_2}{f_1 - f_2} = \frac{(1.0)(0.3)}{1.0 - 0.3} \approx 0.429 \text{ m} \].
05

Determine the Film Plane Distance from the Secondary Mirror

The final image position, considering the displacement of the secondary mirror, is calculated from \(d_i' = d - f \). Thus, \( d_i' = 1.3 - 0.429 = 0.871 \text{ m} \). So, the film plane should be 0.871 m away from the secondary mirror.

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Key Concepts

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

Spherical Mirrors
Spherical mirrors are vital components in optical systems and are based on segments of a sphere. They come in two main forms: concave and convex mirrors.
These mirrors work by reflecting light, not refracting it like lenses do.
Concave mirrors, like those used in reflecting telescopes, have a surface that curves inward like the inside of a bowl.
  • They can converge light to a focal point, making them useful for focusing light rays from distant stars.
  • These mirrors are often used where magnification is important, such as in telescopes and certain cameras.
Conversely, convex mirrors have a surface that bulges outward and are good for diverging light.
Convex mirrors give a wider field of view and are often used in side and rearview mirrors in vehicles.
In the telephoto problem discussed, both mirrors are concave because they need to focus and redirect light.
Focal Length
The focal length of a mirror is definitively important in understanding how the mirror will bend light rays.
It is defined as the distance between the mirror's surface and the point where all parallel incoming light rays converge or appear to diverge from.
The focal length (\( f \)) of a spherical mirror is half its radius of curvature (\( R \)).
  • For the primary mirror with \( R = 2.0 \, \text{m} \), the focal length is \( f_1 = \frac{2.0}{2} = 1.0 \, \text{m} \).
  • For the secondary mirror with \( R = 0.6 \, \text{m} \), it becomes \( f_2 = \frac{0.6}{2} = 0.3 \, \text{m} \).
Understanding these values is crucial for calculating how and where images are formed by these mirrors.
Mirror Equation
The mirror equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) relates the focal length (\( f \)), object distance (\( d_o \)), and image distance (\( d_i \)).
This formula helps determine where an image will form relative to the mirror. In our problem:
  • Because the object (a star) is extremely far away,\( d_o \approx \infty \).
  • This simplifies the equation to \( d_i \approx f_1 \), meaning the image forms approximately at the focal length of the primary mirror.
This understanding is key for determining distances in complex systems involving multiple mirrors.
Effective Focal Length
When dealing with a system of multiple mirrors, the effective focal length can be a bit tricky but crucial for calculating where light finally converges.
  • For two mirrors, like in our exercise, the effective focal length (\( f \) ) is calculated using the formula: \( \frac{1}{f} = \frac{1}{f_1} - \frac{1}{f_2} \).
  • Substitute the known focal lengths: \( f_1 = 1.0 \, \text{m} \) and \( f_2 = 0.3 \, \text{m} \), to get \( f \approx 0.429 \, \text{m} \).
This value indicates where the final image will form, helping us position tools like the film plane in setups such as our telephoto lens system. Understanding effective focal length is pivotal for combining optical elements to control focus and clarity.

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Most popular questions from this chapter

What must the focal length of a thin negative lens be for it to form a virtual image \(50 \mathrm{cm}\) away (measured from the lens) of an ant located \(100 \mathrm{cm}\) away (measured from the lens)? Given (just as a change of pace) that the ant is to the right of the lens, locate and describe its image.

Construct a Cartesian oval such that the conjugate points will be separated by \(11 \mathrm{cm}\) when the object is \(5 \mathrm{cm}\) from the vertex. If \(n_{1}=1\) and \(n_{2}=\frac{3}{2},\) draw several points on the required surface.

Consider the case of two positive thin lenses, \(L_{1}\) and \(L_{2},\) separated by \(5 \mathrm{cm} .\) Their diameters are 6 and \(4 \mathrm{cm},\) respectively, and their focal lengths are \(f_{1}=9 \mathrm{cm}\) and \(f_{2}=3 \mathrm{cm} .\) If a diaphragm with a hole \(1 \mathrm{cm}\) in diameter is located between them, \(2 \mathrm{cm}\) from \(L_{2},\) find (a) the aperture stop and (b) the locations and sizes of the pupils for an axial point, \(S, 12 \mathrm{cm}\) in front of (to the left of) \(L_{1}\).

A thin double-convex glass lens (with an index of 1.56 ) while surrounded by air has a 10 -cm focal length. If it is placed under water (having an index of 1.33 ) \(100\) \(\mathrm{cm}\) beyond a small fish, where will the guppy's image be formed?

A thin, straight piece of wire \(4.00 \mathrm{mm}\) long is located in a plane perpendicular to the optical axis and \(60.0 \mathrm{cm}\) in front of a thin lens. The sharp image of the wire formed on a screen is \(2.00 \mathrm{mm}\) long. What is the focal length of the lens? When the screen is moved farther from the lens by \(10.0 \mathrm{mm}\), the image blurs to a width of \(0.80 \mathrm{mm}\). What is the diameter of the lens? [Hint: Image a source point on the axis.]
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