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34.13 Dental Mirror. 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 1.25cm 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

a) The dentist must use concave mirror.

b) The focal length is 2.5cm and radius of curvature is 5cm.

c) The principal ray diagram is shown below

Step by step solution

01

Lateral magnification and object image relation for spherical mirror

Lateral magnification for spherical message is given by:

m=-vu=hiho

where, m is the magnification,

u is object distance from the vertex of the mirror,

u is (+) when the object is in front of the reflecting surface of mirror,

v is image distance from the vertex of the mirror,

v is (+) when the image is in front of the reflecting surface of mirror,

is the height of the image, is the height of the object,

m is (+) when the image is erect and (-) when the image is inverted

Object image relation for spherical mirror is given by:

1u+1v=2R=1f

Where, u= object distance from the vertex of the mirror,

u is (+) when the object is in front of the reflecting surface of mirror,

v is image distance from the vertex of the mirror,

v is (+) when the image is in front of the reflecting surface of mirror,

R is the radius of curvature of the mirror, f= focal length,

R is (+) in concave mirror and (-) in convex mirror,

u and v are negative if the case is opposite to what is mentioned above

Remember that the focal length of a mirror depends only on the curvature of the mirror and not on the material from which the mirror is made because the formation of the image results from rays reflected from the surface of the material.

Apply: The above rule can be applied on either concave and convex mirrors or even plane mirror when we consider the radius to be infinity. The equation gives one the exact location of an image by knowing the radius or the focal length of a mirror and the location of the object. In addition, the sign for the distance of the image can be negative if the image in the opposite direction of the outgoing rays which means the image is not real image but is virtual one and is located inside the mirror.

02

Determine the kind of mirror needed

a) Only a concave mirror can produce an upright and magnified image. Convex mirror can produce upright image, but it is diminished.

Hence the dentist must use concave mirror to get magnification 2.

03

Determine the focal length and radius of curvature of mirror

b) By lateral magnification,

m=-vuv=-muv=-2.1.25=-2.5cm

Now, using object image relation for spherical mirror,

Now,1u+1v=1f

Solve for f to get

f=uvu+v

Substitutev=-2.5cm andu=1.25cm

f=1.25.-2.51.25-2.5=2.5cm

Positive sign confirms the mirror is concave.

R=2f=2.2.5=5cm

Hence, the focal length is 2.5cm and radius of curvature is 5cm.

04

Draw a principal-ray diagram

The figure below shows the principal ray diagram where image is virtual and magnified by 2.

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Most popular questions from this chapter

A pencil that is 9cm long is held perpendicular to the surface of a plane mirror with the tip of the pencil lead 12cm from the mirror surface and the end of the eraser 21cm from the mirror surface. What is the length of the image of the pencil that is formed by the mirror? Which end of the image is closer to the mirror surface: the tip of the lead or the end of the eraser?

34.14 For a convex spherical mirror that has focal length f=-12cm, what is the distance of an object from the mirrorโ€™s vertex if the height of the image is half the height of the object?

On December 26, 2004, a violent earthquake of magnitude 9.1 occurred off the coast of Sumatra. This quake triggered a huge tsunami (similar to a tidal wave) that killed more than 150,000 people. Scientists observing the wave on the open ocean measured the time between crests to be 1.0 h and the speed of the wave to be 800 km>h. Computer models of the evolution of this enormous wave showed that it bent around the continents and spread to all the oceans of the earth. When the wave reached the gaps between continents, it diffracted between them as through a slit.

(a) What was the wavelength of this tsunami?

(b) The distance between the southern tip of Africa and northern Antarctica is about 4500 km, while the distance between the southern end of Australia and Antarctica is about 3700 km. As an approximation, we can model this waveโ€™s behavior by using Fraunhofer diffraction. Find the smallest angle away from the central maximum for which the waves would cancel after going through each of these continental gaps.

You use a lens of diameter and light of wavelength and frequency to form an image of two closely spaced and distant objects. Which of the following will increase the resolving power? (a) Use a lens with a smaller diameter; (b) use light of higher frequency; (c) use light of longer wavelength. In each case justify your answer.

According to the discussion in Section 34.2, light rays are reversible. Are the formulas in the table in this chapterโ€™s Summary still valid if object and image are interchanged? What does reversibility imply with respect to the forms of the various formulas?

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