/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 An insect \(3.75 \mathrm{~mm}\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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An insect \(3.75 \mathrm{~mm}\) tall is placed \(22.5 \mathrm{~cm}\) to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude \(13.0 \mathrm{~cm},\) and the index of refraction of the lens material is \(1.70 .\) (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Upright or inverted? (b) Repeat part (a) if the lens is reversed.

Short Answer

Expert verified
Image location: 106.4 cm, Size: 17.7 mm, Real, Upright. Reversed lens: Same.

Step by step solution

01

Calculate Lens Focal Length

The focal length for a planoconvex lens can be determined using the lensmaker's equation:\[ \frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]Given:- For a planoconvex lens, left surface is flat, so \( R_1 = \infty \)- Right surface, \( R_2 = -13.0 \text{ cm} \) (negative because it's convex when viewed from the object side)Thus:\[ \frac{1}{f} = (1.70 - 1)\left(\frac{1}{\infty} - \frac{1}{-13.0}\right) \]\[ \frac{1}{f} = 0.70 \left( 0 + \frac{1}{13.0} \right) \]\[ \frac{1}{f} = \frac{0.70}{13.0} \approx 0.0538 \]\[ f = \frac{1}{0.0538} \approx 18.6 \text{ cm} \]
02

Use Lens Formula to Find Image Location

The lens formula is given by:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Given the object distance \( d_o = -22.5 \text{ cm} \), substitute the values:\[ \frac{1}{18.6} = \frac{1}{-22.5} + \frac{1}{d_i} \]Solving for \( d_i \):\[ \frac{1}{d_i} = \frac{1}{18.6} + \frac{1}{22.5} \approx 0.0538 - 0.0444 \]\[ \frac{1}{d_i} \approx 0.0094 \]\[ d_i \approx \frac{1}{0.0094} \approx 106.4 \text{ cm} \]The positive value indicates the image is real and formed on the opposite side of the lens.
03

Calculate Magnification and Image Size

The magnification \( m \) is given by:\[ m = -\frac{d_i}{d_o} \]\[ m = -\frac{106.4}{-22.5} \approx 4.73 \]The image size \( h_i \) is then:\[ h_i = m \times h_o \]Given the object height \( h_o = 3.75 \text{ mm} \):\[ h_i = 4.73 \times 3.75 \approx 17.7 \text{ mm} \]Since the magnification is positive, the image is upright.
04

Analysis When Lens is Reversed

When the lens is reversed, the left surface is now the convex one and the right is flat:The focal length does not change because:\[ \frac{1}{f} = (n - 1) \left( \frac{1}{-R_2} - \frac{1}{\infty} \right) = (1.70 - 1) \left( \frac{1}{13.0} \right) \]So the same focal length applies, \( f = 18.6 \text{ cm} \).Repeating the lens formula calculations confirm the image characteristics of being real, inverted, and of the same size.

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Key Concepts

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

Planoconvex Lens
A planoconvex lens is a type of lens with one flat (plane) surface and one outward-curving (convex) surface. This design is frequently used because it allows for efficient light focusing. Such lenses are often used in applications that require collimated light, like laser systems.
Here are some key points about planoconvex lenses:
  • They converge light to a focal point, ideal for focusing applications.
  • The flat side often faces the light source, minimizing spherical aberration.
  • The curvature is involved in determining the lens's optical power.
Their function in optical systems is crucial. By bending incoming parallel light rays towards a focal point, planoconvex lenses can create real images. This property makes them valuable in any system that requires focused light.
Lensmaker's Equation
The lensmaker's equation is an essential formula in geometric optics. It relates the focal length of a lens to the curvature of its surfaces and the refractive index of the material. This equation is expressed as:\[\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]where:
  • \( f \) is the focal length of the lens.
  • \( n \) is the refractive index of the lens material.
  • \( R_1 \) and \( R_2 \) are the radii of curvature for the lens's two surfaces.
For planoconvex lenses, one radius is infinite because of the flat surface. This simplifies the equation, making calculations easier. The lensmaker's equation allows us to design and evaluate lenses based on desired focal properties. In the given exercise, you apply this formula to find the lens's focal length, which is crucial for determining image formation.
Image Formation
Image formation in lenses results from the bending of light rays as they pass through the lens. A planoconvex lens can form real images when positioned appropriately in relation to the object.
  • Real images are produced on the opposite side of the lens from the object.
  • They can be projected onto a screen because the light actually converges at a physical point.
  • The equations used to determine the image distance include the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) , where \(d_o\) is the object distance and \(d_i\) is the image distance.
In the problem, after employing the lensmaker's equation to find the focal length, the next step was to use this to calculate the image location. The resultant positive image distance indicated that the lens forms a real image, a common attribute of planoconvex lenses under typical orientations. This understanding is vital in various applications, such as imaging systems and optical instrumentation.
Magnification
Magnification is a measure of how much larger or smaller the image is compared to the object itself. This concept is crucial for understanding how optics change the apparent size of objects.
The magnification \( m \) is given by:\[m = -\frac{d_i}{d_o}\]A positive magnification means the image is upright compared to the object, while a negative value indicates an inverted image. It also tells us the size ratio:
  • If \(|m| > 1\), the image is larger than the object.
  • If \(|m| < 1\), the image is smaller than the object.
In the exercise, after calculating the image location, the magnification formula was used to learn that the image is approximately 4.73 times larger than the original object. Additionally, due to the positive magnification value, the planoconvex lens produces an upright image. This concept is pivotal in many fields, including photography and microscopy, where image size and orientation are crucial parameters.

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Most popular questions from this chapter

If you run away from a plane mirror at \(2.40 \mathrm{~m} / \mathrm{s},\) at what speed does your image move away from you?

The two surfaces of a plastic converging lens have equal radii of curvature of \(22.0 \mathrm{~cm},\) and the lens has a focal length of \(20.0 \mathrm{~cm} .\) Calculate the index of refraction of the plastic.

A converging lens with a focal length of \(12.0 \mathrm{~cm}\) forms a virtual image \(8.00 \mathrm{~mm}\) tall, \(17.0 \mathrm{~cm}\) to the right of the lens. Determine the position and size of the object. Is the image upright or inverted? Are the object and image on the same side or opposite sides of the lens?

The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about \(8.0 \mathrm{~mm},\) which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object \(16 \mathrm{~cm}\) tall were placed \(30.0 \mathrm{~cm}\) from the eye lens, where would the lens focus it and how tall would the image be? Is this image real or virtual? Is it upright or inverted? (Note: The results obtained here are not strictly accurate because the lens is embedded in fluids having refractive indices different from that of air.)

The focal length of a mirror can be determined entirely from the shape of the mirror. In contrast, to determine the focal length of a lens, we must know both the shape of the lens and its index of refraction - as well as the index of refraction of the surrounding medium. For instance, when a thin lens is immersed in a liquid, we must modify the thin-lens equation to take into account the refractive properties of the surrounding liquid: $$\frac{1}{f}=\left(\frac{n}{n_{\mathrm{liq}}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right),$$ where \(n_{\text {liq }}\) is the index of refraction of the liquid and \(n\) is the index of refraction of the glass. If you place a concave mirror with a focal length of \(1 \mathrm{~m}\) into water, what will happen? A. The mirror will no longer be able to focus light. B. The focal length of the mirror will decrease. C. The focal length of the mirror will increase. D. Nothing will happen.

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