Ch 34: Ray Optics

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 34: Ray Optics

Problem 65

Chapter 34, Problem 65

A 2.0-cm-tall candle flame is 2.0 m from a wall. You happen to have a lens with a focal length of 32 cm. How many places can you put the lens to form a well-focused image of the candle flame on the wall? For each location, what are the height and orientation of the image?

Verified step by step guidance

Step 1: Start by identifying the lens equation, which relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\): \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). Here, \(f = 32\,\text{cm}\), \(d_o\) is the distance from the candle to the lens, and \(d_i\) is the distance from the lens to the wall. The total distance between the candle and the wall is \(d_o + d_i = 200\,\text{cm}\).

Step 2: Substitute \(d_i = 200 - d_o\) into the lens equation. This gives \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{200 - d_o}\). Substitute \(f = 32\,\text{cm}\) into the equation: \(\frac{1}{32} = \frac{1}{d_o} + \frac{1}{200 - d_o}\).

Step 3: Solve the equation \(\frac{1}{32} = \frac{1}{d_o} + \frac{1}{200 - d_o}\) for \(d_o\). This is a quadratic equation in terms of \(d_o\). Rearrange and simplify to find the possible values of \(d_o\). These values represent the two possible positions of the lens where a focused image can be formed.

Step 4: Once the values of \(d_o\) are determined, calculate the corresponding \(d_i\) using \(d_i = 200 - d_o\). Then, determine the magnification \(M\) using the formula \(M = -\frac{d_i}{d_o}\). The magnification will help determine the height and orientation of the image. The image height \(h_i\) is given by \(h_i = M \cdot h_o\), where \(h_o = 2.0\,\text{cm}\) is the height of the candle flame.

Step 5: Analyze the results. For each lens position, note the image height \(h_i\) and whether the image is upright (positive \(M\)) or inverted (negative \(M\)). Summarize the two possible lens positions, the corresponding image heights, and orientations.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lens Formula

The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens through the equation 1/f = 1/v - 1/u. This formula is essential for determining where to place the lens to form a focused image, as it allows us to calculate the image distance based on the object distance and the lens's focal length.

Magnification

Magnification (M) is the ratio of the height of the image (h') to the height of the object (h), expressed as M = h'/h = -v/u. This concept is crucial for understanding how the size and orientation of the image change based on the position of the lens relative to the candle flame and the wall.

Real and Virtual Images

Real images are formed when light rays converge and can be projected onto a screen, while virtual images occur when light rays diverge and cannot be projected. Understanding the difference between these types of images is important for determining the characteristics of the image formed by the lens at various positions relative to the candle flame and the wall.

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Textbook Question

A 25-cm-long rod lies along the optical axis of a converging lens, perpendicular to the lens plane. The lens has a 30 cm focal length. The rod's real , along the optical axis on the other side of the lens, is also 25 cm long. What is the distance from the lens to the nearest end of the rod?

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Textbook Question

BIO A dentist uses a curved mirror to view the back side of teeth in the upper jaw. Suppose she wants an upright image with a magnification of 1.5 when the mirror is 1.2 cm from a tooth. Should she use a convex or a concave mirror? What focal length should it have?

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Textbook Question

An old-fashioned slide projector needs to create a 98-cm-high of a 2.0-cm-tall slide. The screen is 300 cm from the slide. What focal length does the lens need? Assume that it is a thin lens.

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Textbook Question

Paraxial light rays approach a transparent sphere parallel to an optical axis passing through the center of the sphere. The rays come to a focus on the far surface of the sphere. What is the sphere's index of refraction?

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Textbook Question

A lightbulb is 3.0 m from a wall. What are the focal length and the position (measured from the bulb) of a lens that will form an on the wall that is twice the size of the lightbulb?

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Textbook Question

BIO A keratometer is an optical device used to measure the radius of curvature of the eye's cornea—its entrance surface. This measurement is especially important when fitting contact lenses, which must match the cornea's curvature. Most light incident on the eye is transmitted into the eye, but some light reflects from the cornea, which, due to its curvature, acts like a convex mirror. The keratometer places a small, illuminated ring of known diameter 7.5 cm in front of the eye. The optometrist, using an eyepiece, looks through the center of this ring and sees a small virtual image of the ring that appears to be behind the cornea. The optometrist uses a scale inside the eyepiece to measure the diameter of the image and calculate its magnification. Suppose the optometrist finds that the magnification for one patient is 0.049. What is the absolute value of the radius of curvature of her cornea?

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