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Chapter 5: Problem 46

A blade of grass standing \(10.0 \mathrm{mm}\) tall is \(150 \mathrm{mm}\) in front of a thin positive lens having a \(100 \mathrm{mm}\) focal length; \(250 \mathrm{mm}\) behind that first lens is a thin negative lens with a focal length of \(-75.0 \mathrm{mm}\). (a) Show that the first lens forms an image \(300 \mathrm{mm}\) behind it. (b) Describe that image. (c) What's its magnification? (d) Prove that the final image formed by both lenses is located \(150 \mathrm{mm}\) behind the negative lens. (e) What is the total magnification of the combination?

Short Answer

Expert verified
(a) Image distance is 300 mm. (b) Real and inverted. (c) Magnification is -2. (d) Final image distance is 150 mm. (e) Total magnification is -6.

Step by step solution

01

Calculate Image Distance for First Lens

To find the image distance for the first lens, we use the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f = 100 \mathrm{mm} \) is the focal length of the lens and \( d_o = 150 \mathrm{mm} \) is the object distance. Plugging in these values, we have:\[\frac{1}{100} = \frac{1}{150} + \frac{1}{d_i}\]Solving for \( d_i \), we rearrange the equation:\[\frac{1}{d_i} = \frac{1}{100} - \frac{1}{150} = \frac{3 - 2}{300} = \frac{1}{300}\]Thus, \( d_i = 300 \mathrm{mm} \).
02

Describe the Image Formed by the First Lens

The image formed by the first lens is real and inverted because the image distance \( d_i = 300 \mathrm{mm} \) is positive and greater than the focal length. Since the image is formed at a distance greater than the lens's focal point, it is located on the opposite side of the incoming light.
03

Calculate the Magnification for the First Lens

The magnification \( m \) provided by a lens is given by the formula \( m = -\frac{d_i}{d_o} \). For the first lens:\[m = -\frac{300}{150} = -2\]Thus, the image is magnified by a factor of 2 and inverted.
04

Determine the Image Distance for the Second Lens

The first image is located \( 50 \mathrm{mm} \) in front of the second lens because the initial image is \( 300 \mathrm{mm} \) from the first lens, and the distance between the lenses is \( 250 \mathrm{mm} \). Therefore, the object distance for the second lens is \( d_o = -50 \mathrm{mm} \) (negative because it's on the same side as the incoming light).Using the lens equation for the second lens with \( f = -75 \mathrm{mm} \):\[\frac{1}{-75} = \frac{1}{-50} + \frac{1}{d_i}\]Solving for \( d_i \),\[\frac{1}{d_i} = \frac{1}{-75} - \frac{1}{-50} \frac{1}{d_i} = \frac{-2}{150} + \frac{3}{150} = \frac{1}{150}\]Therefore, \( d_i = 150 \mathrm{mm} \). The image distanceshare the sign convention indicating it is real and located on the opposite side of the second lens.
05

Calculate the Total Magnification of the Lens Combination

The total magnification is the product of the individual magnifications from both lenses. For the first lens, the magnification is \(-2\). The magnification from the second lens can be calculated as \( m_2 = -\frac{d_i}{d_o} = -\frac{150}{-50} = 3 \).Thus, the total magnification \( m_{total} = m_1 \times m_2 = (-2) \times (3) = -6 \). The final image is magnified by a factor of 6 and is inverted.

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Key Concepts

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

Lens Formula
The lens formula is crucial when working with optical lenses. It connects the focal length of a lens (f), the object distance (d_o), and the image distance (d_i) using the equation:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This formula helps us calculate how the lens will refract light to form an image. For instance, in the problem, we used this formula to find the image distance for the first lens with a known object distance and focal length. By rearranging the formula, \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]we solved for d_i.
The solution showed that the image formed by the first lens was300 mmbehind it, confirming the calculation using the lens formula.
Image Formation
When light passes through a lens, it is bent or refracted, creating an image. The nature of this image depends on several factors:
  • The type of lens (convex or concave)
  • The position of the object relative to the lens
  • The focal length of the lens
In the given exercise, the first lens is positive (convex), creating a real image because the calculated image distance is positive and greater than the focal length. This means the light converges on the opposite side of the lens from the object.
The second negative lens (concave) transforms this image into yet another location. Here, the light appears to diverge from a virtual point on the same side as the initial image.
We found the final image to be 150 mm behind the negative lens, showcasing a complex yet fascinating journey light takes through multiple lenses.
Magnification
Magnification describes how much larger or smaller an image is compared to the object. It is calculated using the formula:\[ m = -\frac{d_i}{d_o} \]This negative sign indicates whether the image is inverted. In our problem, the first lens magnified the image by a factor of -2, meaning the image was twice the size and inverted. For the entire lens system, we found the total magnification by multiplying the magnifications from each lens.
The second lens had a positive magnification of 3. Consequently, the overall effect was a total magnification of -6, resulting in an image six times larger than the original object and inverted.
Focal Length
The focal length is a fundamental characteristic of a lens. It is the distance from the lens where parallel rays of light are brought to a focus (for convex lenses) or appear to diverge from (for concave lenses).
In the exercise, we dealt with both positive and negative lenses. The positive lens had a focal length of 100 mm , meaning it converged light to a point that far from the lens.
On the other hand, the negative lens had a focal length of -75 mm , indicating its nature of diverging light. Focal lengths determine how dramatically a lens bends light, thus affecting image formation and magnification characteristics. Understanding focal length helps predict how you'll see an object through the lens.

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Most popular questions from this chapter

A point source \(S\) sitting on the central axis of a positive thin lens is located (to the left) between one and two focal lengths from the lens. A concave spherical mirror is to be positioned to the right of the lens so that the final real image also lies at point \(S\). Where should the mirror be placed? Where should a convex spherical mirror be located to accomplish the same feat?

A candle that is \(6.00 \mathrm{cm}\) tall is standing \(10 \mathrm{cm}\) from a thin concave lens whose focal length is \(-30 \mathrm{cm} .\) Determine the location of the image and describe it in detail. Draw an appropriate ray diagram.

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.]

The focal length of a planar-convex thin lens in air is \(250.0 \mathrm{cm}\) The glass it is made of has an index of \(1.530 .\) Determine the radii of curvature of its surfaces. What would happen to the radii if \(n\) was reduced to \(1.500 ?\)

A small planar mirror is attached to a thin vertical wire so that the mirror is parallel to a wall \(1.0 \mathrm{m}\) away. A horizontal scale is mounted flat on the wall opposite the mirror, whose center is directly opposite the zero mark on the scale. A horizontal laserbeam reflects off the mirror and hits the scale at \(5.0 \mathrm{cm}\) left of zero. The mirror is then rotated through an angle \(\alpha\) and the beam-scale spot of light moves left an additional \(15.0 \mathrm{cm} .\) Find \(\alpha\).
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