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32 through 38 37, 38 33, 35 Spherical refracting surfaces. An object Ostands on the central axis of a sphericalrefractingsurface. For this situation, each problem in Table 34-5 refers to the index of refractionn1where the object is located, (a) the index of refraction n2on the other side of the refracting surface, (b) the object distance p, (c) the radius of curvature rof the surface, and (d) the image distance i. (All distances are in centimeters.) Fill in the missing information, including whether the image is (e) real (R)or virtual (V)and (f) on the same side of the surface as object Oor on the opposite side

Short Answer

Expert verified

a) The index of refraction n2on the other side of the refracting surface is 1.0.

b) The object distance p is +10 cm.

c) The radius of curvature r of the surface is 30 cm.

d) The image distance i is -6 cm.

e) The image is virtual and upright.

f) The image is on same side as that of the object.

Step by step solution

01

Step 1: Given

n1=1.5n2=1.0p=+10.0cmi=-6.0

Table 34-5

02

Determining the concept

The index of refraction of object and image, the object distance and the image distance are given i the problem. Using this data and equation, find the radius of curvature and check whether the image is real or virtual and find the position of the image.

Formulae are as follows:

np+n2i=n2-nT

where p is the pole and i is the image distance

03

Determining the index of refraction  on the other side of the refracting surface

(a)

Index of refraction on the other side of the refracting surface is given in the table 34-5. So n2=1.0.

Therefore, the index of refraction n2on the other side of the refracting surface is 1.0.

04

Determining the object distance 

(b)

The index of refraction on the other side of the refracting surface is given in the table 34-5. So, n2=1.0.

Therefore, the object distance p is +10 cm.

05

Determining the radius of curvature r of the surface

(c)

Theobject distance is given in the problem, p = +10 cm

Therefore, the radius of curvature r of the surface is 30 cm.

06

Determining the image distance i

d)

From equation 34-8

n1p+n2i=n2-n1r

Rearranging the terms,

r=n2-nnp+n2i

Substituting the given values

role="math" localid="1663044974048" r=1.0-1.51.510+1.06r=30cm

Therefore, the image distance i is -6 cm.

07

Determining whether the image is real or virtual

(e)

The image distance is given in the problem i = -6 cm

Therefore, the image is virtual and upright.

08

Determining the position of the image

(f)

For spherical refracting surfaces, real images form on the opposite side of the object and virtual images form on the same side as the object.

Since the image is virtual, therefore theimage is on the same side as that of the object.

Therefore, the image is on same side as that of the object.

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

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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.)

In Fig. 34-54, a fish watcher at point P watches a fish through a glass wall of a fish tank. The watcher is level with the fish; the index of refraction of the glass is 8/5, and that of the water is 4/3. The distances are d1=8.0cm,d2=3.0cm,d3=6.8cm. (a) To the fish, how far away does the watcher appear to be? (Hint: The watcher is the object. Light from that object passes through the walls outside surface, which acts as a refracting surface. Find the image produced by that surface. Then treat that image as an object whose light passes through the walls inside surface, which acts as another refracting surface.) (b) To the watcher, how far away does the fish appear to be?

An object is moved along the central axis of a thin lens while the lateral magnification m is measured. Figure 34-43 gives m versus object distance p out to ps. What is the magnification of the object when the object is p=14 cmfrom the lens?

In Fig. 34-52, an object is placed in front of a converging lens at a distance equal to twice the focal length f1of the lens. On the other side of the lens is a concave mirror of focal lengthf2separated from the lens by a distance 2(f1+f2). Light from the object passes rightward through the lens, reflects from the mirror, passes leftward through the lens, and forms a final image of the object. What are (a) the distance between the lens and that final image and (b) the overall lateral magnification M of the object? Is the image (c) real or virtual (if it is virtual, it requires someone looking through the lens toward the mirror), (d) to the left or right of the lens, and (e) inverted or non-inverted relative to the object?

(a) A luminous point is moving at speedV0toward a spherical mirror with a radius of curvaturer, along the central axis of the mirror. Show that the image of this point is moving at the speed

vI=-(r2p-r)2v0

Where,p is the distance of the luminous point from the mirror at any given time. Now assume the mirror is concave, withr=15cm.and letV0=5cm/s. FindV1when (b)p=30cm(far outside the focal point), (c) p=8.0cm(just outside the focal point), and (d)p=10mm(very near the mirror).

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