In Fig. 34-38, a beam of parallel light rays from a laser is incident on a solid transparent sphere of an index of refraction n. (a) If a point image is produced at the back of the sphere, what is the index of refraction of the sphere? (b) What index of refraction, if any, will produce a point image at the center of the sphere?

Short Answer

Expert verified
  1. If a point image is produced at the back of the sphere, the refractive index of the sphere is 2.00.
  2. It is not possible to produce a point image at the center of the sphere.

Step by step solution

01

Step 1: Given data

  • The object distance isp=
  • Refractive index;n1=nair=1
02

Determining the concept

Using the relation between the index of refraction of object and image, image distance, object distance, and the radius of curvature, given by equation 34-8, find the required answers.

Formulae are as follows:

n1p+n2i=n2-n1r

Here, p is the pole, i is the image distance.

03

(a) Determining the refractive index of the sphere, if a point image is produced at the back of the sphere.

As the image is produced at the back of the sphere, so,

i = 2r

n1p+n2i=n2-n1r

Substituting the given values,

1+n22r=n2-1r

n22r=n2-1rn22=n2-1n21-12=1n22=1n2=2.00

Hence, if a point image is produced at the back of the sphere, the refractive index of the sphere is 2.00.

04

(b) Determine the refractive index of the sphere that will produce a point image at the center of the sphere.

For an image to be produced at the center of the sphere,

i=rn1p+n2i=n2-n1r

Substituting the given values,

role="math" localid="1662977840109" n1+n2r=n2-1rn2r=n2-1r

This is not valid unless n2or r

Hence, it is impossible to produce a point image at the center of the sphere.

The required quantities can be found by using the relation between the index of refraction of object and image, the image distance, the object distance, and the radius of curvature.

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Most popular questions from this chapter

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’s radius of curvature. Next, use tanβ=yxto find the angle b at the point of intersection, and then useα+y=2βtofind 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 — 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.

80 through 87 80, 87 SSM WWW 83 Two-lens systems. In Fig. 34-45, stick figure O (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to O, which is at object distance p1. Lens 2 is mounted within the farther boxed region, at distance d. Each problem in Table 34-9 refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by C for converging and D for diverging; the number after C or D is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance i2for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification Mfor the system, including signs.Also, determine whether the final image is (c) real(R) or virtual (V), (d) inverted (I)from object O or non-inverted (NI), and (e) on the same side of lens 2 as the object O or on the opposite side.

A peanut is placed 40cmin front of a two-lens system: lens 1 (nearer the peanut) has focal length f1 =20cm, lens 2 has f2=-15cm and the lens separation is d=10cm. For the image produced by lens 2, what are (a) the image distance i2(including sign), (b) the image orientation (inverted relative to the peanut or not inverted), and (c) the image type (real or virtual)? (d) What is the net lateral magnification?

In Fig. 34-26, stick figure O stands in front of a thin, symmetric lens that is mounted within the boxed region; the central axis through the lens is shown. The four stick figuresI1andI4suggest general locations and orientations for the images that might be produced by the lens. (The figures are only sketched in; neither their height nor their distance from the lens is drawn to scale.) (a) Which of the stick figures could not possibly represent images? Of the possible images, (b) which would be due to a converging lens, (c) which would be due to a diverging lens, (d) which would be virtual, and (e) which would involve negative magnification?

69 through 79 76, 78 75, 77 More lenses. Objectstands on the central axis of a thin symmetric lens. For this situation, each problem in Table 34-8 refers to (a) the lens type, converging or diverging , (b) the focal distance , (c) the object distance , (d) the image distance , and (e) the lateral magnification . (All distances are in centimetres.) It also refers to whether (f) the image is real or virtual , (g) inverted (I)or non-inverted(NI) from , and (h) on the same side of the lens asor on the opposite side. Fill in the missing information, including the value of m when only an inequality is given, where only a sign is missing, answer with the sign.

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