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A small cup of green tea is positioned on the central axis of a spherical mirror. The lateral magnification of the cup is +0.250, and the distance between the mirror and its focal point is 2.00cm. (a) What is the distance between the mirror and the image it produces? (b) Is the focal length positive or negative? (c) Is the image real or virtual?

Short Answer

Expert verified
  1. Image distance from the mirror i = -1.50 cm.
  2. Focal lengthis negative.
  3. The image is virtual.

Step by step solution

01

Listing the given quantities

Magnification, m = +0.250

Focal length, f = 2.0cm.

02

Understanding the concepts of magnification

From the magnification value, we can determine the type of mirror. Then we can use the mirror formula to find the object distance from the mirror and magnification relation to relate image distance and object distance.

Formula:

Mirror formula 1f=1p+1i

Magnification, m=-ip

03

 Step 3: Calculation of the image distance from the mirror

(a)

We have the lateral magnification of cup.

Magnification, m= +0.250

As magnification is positive, the mirror is convex type. Hence,thefocal length is negative.

Now usingthemirror formula, we have

1p-10.250p=1f

1p-4p=1f

This equation gives

-3p=1f=1-2.0=0.5cm-1p=30.5cm

=6.0cm

Using magnification equation,

i= -mp

= -0.250 * 6.0cm

= - 1.50cm

04

Focal length

(b)

We have from part (a) that the focal length is negative.

It is f = -2.0cm.

05

Type of the image

(c)

As the focal length is negative and the mirror is convex, the image is virtual.

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Most popular questions from this chapter

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?

An object is moved along the central axis of a spherical mirror while the lateral magnification m of it is measured. Figure 34-35 gives m versus object distance p for the rangepa=2cm and pb=8.0cm. What is m for p=14cm?

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.

An object is 20cmto the left of a thin diverging lens that has a 30cmfocal length. (a) What is the image distance i? (b) Draw a ray diagram showing the image position.

A double-convex lens is to be made of glass with an index of refraction of 1.5.One surface is to have twice the radius of curvature of the other and the focal length is to be 60mm. What is the (a) smaller and (b) larger radius?

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