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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 (centimeters), the type of mirror, and then the distance p(centimeters, without proper sign) between the focal point and the mirror. Find (a) the radius of curvature r(including sign), (b) the image distance i, and (c) the lateral magnification m. 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 is 70cm.

(b) The image distance is 13.5cm.

(c) Lateral magnification is +0.61.

(d) The image is virtual V.

(e) The image is non-inverted NI.

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

Step by step solution

01

The given data:

The focal length of the mirror, f=35cm

The object distance from the mirror, p=+22cm

The mirror is convex.

02

The concept of the properties of a convex mirror:

A convex mirror or diverging mirror is a curved mirror in which the reflecting surface bulges towards the light source.

The focal length is positive if the mirror is a concave mirror. The focal length is negative if the mirror is a convex mirror. The image distance is positive if the image is a real image and is on the mirror side of the object.

Magnification refers to the ratio of image length to object length measured in planes that are perpendicular to the optical axis.

Virtual, upright and reduced images are always formed by convex mirrors, regardless of the distance between the object and the mirror.

Formulae:

The radius of curvature of a mirror,

r=2f ….. (i)

The mirror equation is,

1f=1i+1p ….. (i)

The lateral magnification of an object,

m=hiho=ip ….. (iii)

Here, fis the focal length of the mirror, pis the object distance from the mirror, iis the image distance, hiis the height of the image, and hois the height of an object.

03

(a) Determining the radius of curvature r:

Since the mirror is convex, the focal length must be negative, i.e.,f=35cm.

Thus, the radius of curvature of the mirror can be given using equation (i) as follows:

r=2f=(2×(35))cm=70cm

Hence, the radius of curvature is 70cm.

04

(b) Determining the image distance i:

Now, the image distance can be calculated by rearranging equation (ii) as follows:

1i=1f1p=pfpf

i=pf(pf)=22cm×(35cm)22cm(35cm)=13.5cm

Therefore, the image distance is 13.5cm.

05

(c) Determining the lateral magnification m:

Thus, the lateral magnification of the mirror can be given using equation (iii) as follows:

m=ip=(13.5cm)22.0cm=+0.61

Therefore, the lateral magnification is +0.61.

06

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

As per the calculations done in part (b), the image distance is found to be negative. This implies that the image formed is opposite to the object placed in front of the mirror (as object distance is positive), thus the image is virtual in nature.

Therefore, the image is virtual V.

07

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

The lateral magnification of the mirror is found to be a positive value as per the calculations based in part (c). Again, you know that the lateral magnification can be given as:

m=hiho=ip=+0.61

Thus, the image height needs to be positive that is possible only in an non-inverted image case.

Hence, 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 of the object O.

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Most popular questions from this chapter

When a T. rex pursues a jeep in the movie Jurassic Park, we see a reflected image of the T. rex via a side-view mirror, on which is printed the (then darkly humorous) warning: “Objects in mirror are closer than they appear.” Is the mirror flat, convex, or concave?

You grind the lenses shown in Fig. 34-53 from flat glass disks (n=1.5)using a machine that can grind a radius of curvature of either 40cmor 60cm. In a lens where either radius is appropriate, you select the 40cmradius. Then you hold each lens in sunshine to form an image of the Sun. What are the (a) focal length fand (b) image type (real or virtual) for (bi-convex) lens 1, (c)f and (d) image type for (plane-convex) lens 2, (e) f and (f) image type for (meniscus convex) lens 3, (g) f and (h) image type for (bi-concave) lens 4, (i) fand (j) image type for (plane-concave) lens 5, and (k) f and (l) image type for (meniscus concave) lens 6?

a real inverted imageof an object is formed by a particular lens (not shown); the object–image separation is, measured along the central axis of the lens. The image is just half the size of the object. (a) What kind of lens must be used to produce this image? (b) How far from the object must the lens be placed? (c) What is the focal length of the lens?

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. 34-60, a sand grain is 3.00cmfrom thin lens 1, on the central axis through the two symmetric lenses. The distance between focal point and lens is 4.00cmfor both lenses; the lenses are separated by 8.00cm. (a) What is the distance between lens 2 and the image it produces of the sand grain? Is that image (b) to the left or right of lens 2, (c) real or virtual, and (d) inverted relative to the sand grain or not inverted?

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