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A point object is 10cmaway from a plane mirror, and the eye of an observer (with pupil diameter5.0mm) is 20cmaway. Assuming the eye and the object to be on the same line perpendicular to the mirror surface, find the area of the mirror used in observing the reflection of the point.

Short Answer

Expert verified

The area of the mirror used in observing the reflection of the point is2.2mm2.

Step by step solution

01

The given data

  1. Object distance,p=10cm
  2. Distance of observer,dey=20cm
  3. Diameter of the pupil,D=5mm
02

Understanding the concept of the geometry

The area of the mirror can be determined by using the object's distance.

Formula:

The area of the circle,A=ฯ€d24,d=diameterofthecircle (i)

03

Calculation of the area of the mirror

Consider the eye of an observer and the object to be in the same line perpendicular to the mirror surface. Consider the two light rays, rand localid="1663053204988" rr, which are closest to and on either sides of the normal ray. Each of these rays has an angle of incidence equal to ฮธwhen they reach the mirror. Consider that these two rays strike the top and bottom edge of the pupil after they have reflected. If ray r strikes the mirror at point Aand ray r'strikes the mirror at, the distance between Aand B, say Xis given as follows:

x=2d0tanฮธ............(a)

Where,d0=pis the distance from the mirror to the object.

We can construct a right triangle starting with the image point of the objectd0. One side of the triangle follows the extended normal axis, and the hypotenuse is along the extension of raylocalid="1663052236991" r. The distance from pupil to Iis,localid="1663052334127" deye+d0and the small angle in this triangle is againฮธ. Thus,

tanฮธ=Rdey+d0โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆb

Where, Ris the pupil radius2.5mm.

Substituting equation (b) in equation (a) with given data, we get the distance between A and B as follows:

x=2d0Rdey+d0=2ร—100mmร—2.5mm200mm+100mm=1.67mm

Now, x serves as the diameter of a circular area A on the mirror, in which all rays that reflect will reach the eye. Thus, the area of the circular area is given using the data in equation (i) as follows:

localid="1663053097431" A=14ร—ฯ€ร—(1.67mm)2=2.2mm2

Hence, the value of the area is2.2mm2.

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Most popular questions from this chapter

An object is placed against the center of a concave mirror and then moved along the central axis until it is 5.0 m from the mirror. During the motion, the distance |i|between the mirror and the image it produces is measured. The procedure is then repeated with a convex mirror and a plane mirror. Figure 34-28 gives the results versus object distance p. Which curve corresponds to which mirror? (Curve 1 has two segments.)

Figure 34-47a shows the basic structure of the human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Fig. 34-47b). A โ€œnormalโ€ eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where the processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Fig. 34-47c). (a) Suppose that for the parallel rays of Figs. 34-47a and b, the focal length fโ€™of the effective thin lens of the eye is 2.50 cm. For an object at distance p = 40 cm, what focal length fโ€™ of the effective lens is required for the object to be seen clearly? (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length fโ€™?

Figure 34-33 shows an overhead view of a corridor with a plane mirror Mmounted at one end. A burglar Bsneaks along the corridor directly toward the center of the mirror. Ifd=3m, how far from the mirror will she from the mirror when the security guardScan first see her in the mirror?

a real inverted imageof an object is formed by a particular lens (not shown); the objectโ€“image separation is, measured along the central axis of the lens. The image is just half the size of the object. (a) What kind of lens must be used to produce this image? (b) How far from the object must the lens be placed? (c) What is the focal length of the lens?

A fruit fly of height H sits in front of lens 1 on the central axis through the lens. The lens forms an image of the fly at a distance d=20cmfrom the fly; the image has the flyโ€™s orientation and height H1=2.0H. What are (a) the focal lengthf1 of the lens and (b) the object distance p1of the fly? The fly then leaves lens 1 and sits in front of lens 2, which also forms an image at d=20cmthat has the same orientation as the fly, but now H1=0.50H. What are (c) f2and (d) p2?

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