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Chapter 34: Q94P (page 1042)

An object is placed against the center of a spherical mirror and then moved 70 cm from it along the central axis as the image distance i is measured. Figure 34-48 gives i versus object distance p out to ps=40cm. What is the image distance when the object is 70 cm from the mirror?

Short Answer

Expert verified

The image distance is inew=-21cm.

Step by step solution

01

Step 1: Given data

  • Object distance,pgiven=70cm .
  • Horizontal Scale, ps=40cm.
02

Determining the concept

The focal length of the lens is constant for every situation. Using this concept, find the focal length from the graph. Using the value in the equation and the given object distance, find the new image distance.

Formulae are as follows:

1f=1p+1i

Here, p is the pole, f is the focal length, and i is the image distance.

03

Determining the image distance

From the graph, take one point Q having co-ordinate Q15,-10.

So,

1p+1i=1f

115+1-10=1f=1pgiven+1inew

Rearrange for the inew;

1inew=-10+15-10脳15-170=-5150-170

Solve for inew;

1inew=-1002100=-21cm

Therefore, the image distance isinew=-21cm .

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Most popular questions from this chapter

A corner reflector, much used in the optical, microwave, and other applications, consists of three plane mirrors fastened together to form the corner of a cube. Show that after three reflections, an incident ray is returned with its direction exactly reversed.

A pepper seed is placed in front of a lens. The lateral magnification of the seed is +0.300. The absolute value of the lens鈥檚 focal length is40.0cm. How far from the lens is the image?

58 through 67 61 59 Lenses with given radii. An object Ostands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance p, index of refraction n of the lens, radius r1of the nearer lens surface, and radius r2of the farther lens surface. (All distances are in centimeters.) Find (a) the image distance and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real (R)or virtual (V), (d) inverted (I)from the object Oor non-inverted (NI), and (e) on the same side of the lens as object Oor on the opposite side

The equation 1p+1i=2rfor spherical mirrors is an approximation that is valid if the image is formed by rays that make only small angles with the central axis. In reality, many of the angles are large, which smears the image a little. You can determine how much. Refer to Fig. 34-22 and consider a ray that leaves a point source (the object) on the central axis and that makes an angle a with that axis. First, find the point of intersection of the ray with the mirror. If the coordinates of this intersection point are x and y and the origin is placed at the center of curvature, then y=(x+p-r)tan a and x2+ y2= r2where pis the object distance and r is the mirror鈥檚 radius of curvature. Next, use tan尾=yxto find the angle b at the point of intersection, and then use伪+y=2尾tofind the value of g. Finally, use the relationtany=y(x+i-r)to find the distance iof the image. (a) Suppose r=12cmand r=12cm. For each of the following values of a, find the position of the image 鈥� that is, the position of the point where the reflected ray crosses the central axis:(0.500,0.100,0.0100rad). Compare the results with those obtained with theequation1p+1i=2r.(b) Repeat the calculations for p=4.00cm.

A concave mirror has a radius of curvature of 24cm. How far is an object from the mirror if the image formed is (a) virtual and 3.0 times the size of the object, (b) real and 3.0 times the size of the object, and (c) real and 1/3 the size of the object?
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