/*! 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 13 A dentist uses a curved mirror t... [FREE SOLUTION] | 91Ó°ÊÓ

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A dentist uses a curved mirror to view teeth on the upper side of the mouth. Suppose she wants an erect image with a magnification of 2.00 when the mirror is \(1.25 \mathrm{~cm}\) from a tooth. (Treat this problem as though the object and image lie along a straight line.) (a) What kind of mirror (concave or convex) is needed? Use a ray diagram to decide, without performing any calculations. (b) What must be the focal length and radius of curvature of this mirror? (c) Draw a principal-ray diagram to check your answer in part (b).

Short Answer

Expert verified
The dentist uses a convex mirror. The focal length of mirror should be \(0.83 cm\). The radius of curvature of the mirror should be \(1.66 cm\).

Step by step solution

01

Kind of mirror

Since the dentist needs an erect image (which means the image formed is upright), a convex mirror is being used. A convex mirror is capable of producing erect image regardless of object's position.
02

Calcualtion of Focal Length

For a mirror, the magnification \('m'\) is given by the formula -\('m = - i / o'\) where 'i' is image distance and 'o' is object distance. Since 'm' is given as 2.00 and the image is erect (which means the image distance 'i' is positive), the object distance 'o' is \(- 1.25 cm\). The negative sign indicates that the object is on the same side of the mirror. So, 'i' is \(-2 \times o\) or \(-2 \times -1.25 = 2.5 cm\). The mirror formula is \(\frac{1}{f} = \frac{1}{i} + \frac{1}{o}\), where 'f' is the focal length. Substituting the values -\(o = -1.25 cm\) and \(i = 2.5 cm\), we get \(f = 0.83 cm\). This focal length is positive signifying the mirror being convex.
03

Calculation of Radius of curvature

For a spherical mirror, focal length f is half of radius of curvature R (i.e., \(f = \frac{R}{2}\)). Substituting the value of f, we find \(R = 2 \times f = 2 \times 0.83 cm = 1.66 cm\) . The radius of curvature is also positive, again signifying the mirror is convex.
04

Drawing Principal-ray Diagram

For checking the answers, draw a principal-ray diagram. The object is placed at the center of curvature (C). One ray of light coming parallel to principal axis, after reflection passes through F (focus). Another ray of light going through F becomes parallel to principal axis after reflection, and both the reflected rays meet at the image position.

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Key Concepts

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

Understanding Mirrors
Mirrors play a crucial role in optics by reflecting light to create images. A mirror can be flat or curved. Curved mirrors are categorized into concave and convex mirrors. Concave mirrors have surfaces that curve inward like a cave, and they can create magnified, reduced, or inverted images based on the object's distance from the mirror. Convex mirrors have surfaces that curve outward. These mirrors always form diminished, upright, and virtual images, which is why they’re often used in applications where a wide field of view is beneficial, like in vehicle side mirrors. Convex mirrors are particularly essential to situations like in the dental practice example, where an erect image is needed.
Focal Length in Mirrors
The focal length of a mirror is a measure of how strongly it converges or diverges light. It is the distance from the mirror to its focal point, where parallel rays of light either converge for concave mirrors or appear to diverge from for convex mirrors. In mathematical terms, the focal length ((f) ) is expressed in the mirror equation: \( \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \), where \( o \) is object distance and \( i \) is image distance. A positive focal length indicates a convex mirror while a negative one indicates a concave mirror. In the exercise, the given convex mirror has a positive focal length, further proving its capability to provide an upright and virtual image as required by the dentist.
Exploring Magnification
Magnification is the ratio of the height of an image to the height of the object and gives us an idea of how large or small the image will appear compared to the actual object. The magnification formula is given by \( m = \frac{i}{o} \), where \( i \) is the image distance and \( o \) is the object distance. For mirrors, if the magnification is positive, the image is upright and virtual, as with convex mirrors. If the magnification is negative, the image is real and inverted, typically a result of using concave mirrors. In the given scenario, a magnification of 2 indicates the image is twice the size of the object. Here, since the magnification is positive, we know the image is upright, a characteristic facilitated by the convex mirror.
Demystifying Convex Mirrors
Convex mirrors, with their outward-curved surface, are vital in situations requiring wide-angle views. They reflect light outwards, which leads to the formation of virtual and diminished images at a location that cannot be projected on a screen, perceived as if they originate from a point behind the mirror. This makes them useful for security and around corners in traffic scenarios. They never produce real images. For the dentist's application where observing an erect image is crucial, the convex mirror is perfect. Due to its configuration, it provides a wide area to be viewed at a glance and ensures images remain upright, which enhances the dentist's ability to see clearly even in hard-to-reach spots.

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Most popular questions from this chapter

When a camera is focused, the lens is moved away from or toward the digital image sensor. If you take a picture of your friend, who is standing \(3.90 \mathrm{~m}\) from the lens, using a camera with a lens with an \(85 \mathrm{~mm}\) focal length, how far from the sensor is the lens? Will the whole image of your friend, who is \(175 \mathrm{~cm}\) tall, fit on a sensor that is \(24 \mathrm{~mm} \times 36 \mathrm{~mm} ?\)

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? Erect or inverted? (b) Repeat part (a) if the lens is reversed.

A tank whose bottom is a mirror is filled with water to a depth of \(20.0 \mathrm{~cm}\). A small fish floats motionless \(7.0 \mathrm{~cm}\) under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

A telescope is constructed from two lenses with focal lengths of \(95.0 \mathrm{~cm}\) and \(15.0 \mathrm{~cm},\) the \(95.0 \mathrm{~cm}\) lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building \(60.0 \mathrm{~m}\) tall, \(3.00 \mathrm{~km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall \(6.00 \mathrm{~m}\) to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?

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