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9, 11, 13 Spherical mirrors. Object O stands on the central axis of a spherical mirror. For this situation, each problem in Table 34-3 gives object distance ps(centimeters), the type of mirror, and then the distance (centimeters, without proper sign) between the focal point and the mirror. Find (a) the radius of curvature r(including sign), (b) the image distancei, and (c) the lateral magnificationm . Also, determine whether the image is (d) real(R)or virtual (V), (e) inverted(I)from object O or non-inverted (NI), and (f) on the same side of the mirror as O or on the opposite side.

Short Answer

Expert verified

(a) The radius of curvature isr=+36cm

(b) The image distance is i=-cm.

(c) Lateral magnification ism=+3.0.

(d) The image is virtual V.

(e) The image is non-invertedNI .

(f) The image is on the opposite side of the object.

Step by step solution

01

Step 1: Given data:

Object distance is,p=+12cm

Focal length of mirror,f=18cm

The mirror is concave.

02

Determining the concept:

The object distance, type of the mirror, and focal length is given in the problem. First, find the radius of curvature from the focal length. Then by using the mirror formula, find the image distance. Use the formula for magnification to find the lateralmagnification. Using these quantities, determine whether the image is real or virtual and inverted or non-inverted. Also, find the position of the image.

Formulae:

The relation between the radius of curvature and focal length of the mirror is,

r=2f.

The spherical mirror equation is given by,

1f=1p+1i

The magnification is define by the following expression.

m=-ip

Where, mis the magnification, p is the pole, fis the focal length.

03

(a) Determining the radius of curvature r :

Use the following formula to find the radius of curvature:

r=2ร—f

Since the mirror is concave, the focal length must be positive, i.e., f=+18cm

r=2ร—18cm=+36cm

Hence, the radius of curvature is +36cm.

04

(b) Determining the image distance i : 

Write the spherical mirror equation as below.

1f=1p+1i

Rearranging the above equation foras below.

1i=1f-1p=p-fpfi=pfp-f

Plugging the known values in the above equation, you get

i=12cmร—18cm12-18cm=-36cm

Therefore, the image distanceis-36cm .

05

(c) The lateral magnification :

Determine the lateral magnification as below.

m=-ip

Substitute role="math" localid="1663072917852" -36cmfor iand 12cmfor pin the above equation, and you obtain

m=--36cm12cm=+3.0

Hence, the lateral magnification is+3.0.

06

(d) Determining whether the image is real or virtual:

Since the image distance is negative, the image is virtualV . The image is formed on the other side of the mirror.

07

(e) Determining whether the image is inverted or non-inverted:

As the magnification is positive. Thus, the image is non-inverted(NI).

08

(f) Determining the position of the image:

For spherical mirrors, real images form on the side of the mirror where the object is located and virtual images form on the opposite side. Since the image is virtual, it is formed on the opposite side as the object.

Hence, the image is on the opposite side as the object.

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Most popular questions from this chapter

A20-mm-thicklayerofwater(n=1.33) floats on a40-mmlocalid="1662979231067" thicklayerofcarbontetrachloridelocalid="1662979325107" (n=1.46)in a tank. A coin lies at the bottom of the tank. At what depth below the top water surface do you perceive the coin? (Hint: Use the result and assumptions of Problem 112 and work with a ray diagram.)

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?

Figure 34-25 shows a fish and a fish stalker in water. (a) Does the stalker see the fish in the general region of point a or point b? (b) Does the fish see the (wild) eyes of the stalker in the general region of point c or point d?

A short straight object of lengthLlies along the central axis of a spherical mirror, a distance pfrom the mirror. (a) Show that its image in the mirror has alength, L'=L(f/(p-f))2(Hint: Locate the two ends of the object.) (b) Show that the longitudinal magnification is equal tom'=(L'/L) is equal to m2, where m is the 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?

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