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All three rays appear to originate from the same point after being reflected, locating the upright virtual image behind the mirror and showing it to be larger than the object. (b) Makeup mirrors are perhaps the most common use of a concave mirror to produce a larger, upright image.

A convex mirror is a diverging mirror ( f size 12{f} {} is negative) and forms only one type of image. It is a case 3 image—one that is upright and smaller than the object, just as for diverging lenses. [link] (a) uses ray tracing to illustrate the location and size of the case 3 image for mirrors. Since the image is behind the mirror, it cannot be projected and is thus a virtual image. It is also seen to be smaller than the object.

Figure (a) shows three incident rays, 1, 2, and 3, falling on a convex mirror. Ray 1 falls parallel, ray 2 falls making an angle with the axis, and ray 3 falls obliquely. These rays after reflection appear to come from a point above the axis. The image is erect and diminished and falls above the axis behind the mirror. Here, the distance from the center of the mirror to focal point F is the focal length small f behind the mirror; the distances of the object and the image from the mirror are d sub o and d sub I, respectively. The heights of the object and the image are h sub o and h sub I, respectively. Figure (b) shows an image of a apparel and clothing show room as viewed in a convex mirror; the image appears to be small in size.
Case 3 images for mirrors are formed by any convex mirror. Ray 1 approaches parallel to the axis, ray 2 strikes the center of the mirror, and ray 3 approaches toward the focal point. All three rays appear to originate from the same point after being reflected, locating the upright virtual image behind the mirror and showing it to be smaller than the object. (b) Security mirrors are convex, producing a smaller, upright image. Because the image is smaller, a larger area is imaged compared to what would be observed for a flat mirror (and hence security is improved). (credit: Laura D’Alessandro, Flickr)

Image in a convex mirror

A keratometer is a device used to measure the curvature of the cornea, particularly for fitting contact lenses. Light is reflected from the cornea, which acts like a convex mirror, and the keratometer measures the magnification of the image. The smaller the magnification, the smaller the radius of curvature of the cornea. If the light source is 12.0 cm from the cornea and the image’s magnification is 0.0320, what is the cornea’s radius of curvature?


If we can find the focal length of the convex mirror formed by the cornea, we can find its radius of curvature (the radius of curvature is twice the focal length of a spherical mirror). We are given that the object distance is d o = 12.0 cm and that m = 0.0320 . We first solve for the image distance d i , and then for f size 12{f} {} .


m = –d i / d o . Solving this expression for d i gives

d i = md o .

Entering known values yields

d i = 0 . 0320 12.0 cm = –0.384 cm. size 12{d rSub { size 8{i} } "=-" left (0 "." "0320" right ) left ("12" "." 0" cm" right )"=-"0 "." "384"" cm"} {}
1 f = 1 d o + 1 d i size 12{ { {1} over {f} } = { {1} over {d rSub { size 8{o} } } } + { {1} over {d rSub { size 8{i} } } } } {}

Substituting known values,

1 f = 1 12.0 cm + 1 0 . 384 cm = 2 . 52 cm . size 12{ { {1} over {f} } = { {1} over {"12" "." 0" cm"} } + { {1} over {-0 "." "384"" cm"} } = { {-2 "." "52"} over {"cm"} } } {}

This must be inverted to find f size 12{f} {} :

f = cm 2 . 52 = –0 . 400 cm . size 12{f= { {"cm"} over { +- 2 "." "52"} } "=-"0 "." "400"" cm"} {}

The radius of curvature is twice the focal length, so that

R = 2 f = 0 . 800 cm. size 12{R=2 lline f rline =0 "." "800"" cm"} {}


Although the focal length f size 12{f} {} of a convex mirror is defined to be negative, we take the absolute value to give us a positive value for R size 12{R} {} . The radius of curvature found here is reasonable for a cornea. The distance from cornea to retina in an adult eye is about 2.0 cm. In practice, many corneas are not spherical, complicating the job of fitting contact lenses. Note that the image distance here is negative, consistent with the fact that the image is behind the mirror, where it cannot be projected. In this section’s Problems and Exercises, you will show that for a fixed object distance, the smaller the radius of curvature, the smaller the magnification.

Questions & Answers

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I have a exam on 12 february
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Practice Key Terms 3

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