58 through 67 61 59 Lenses with given radii. An object Ostands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance O, index of refraction n of the lens, radius of the nearer lens surface, and radius of the farther lens surface. (All distances are in centimeters.) Find (a) the image distance and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real or virtual , (d) inverted from the object Oor non-inverted , and (e) on the same side of the lens as object or on the opposite side.

Short Answer

Expert verified
  1. The image distance is -15cm.
  2. The lateral magnification of the object is+1.5 .
  3. The image is virtual V.
  4. The image is not inverted NIfrom object.
  5. The image on the same side of the object.

Step by step solution

01

The data given

  • The object distance,P=+10cm
  • The index of refraction of the lens,n=1.50
  • The radius of the nearer lens surface,r1=+30cm
  • The radius of the farther lens surface,r2=-30cm
02

Understanding the concept of properties of the lens

We can use the concept of the lens formula and the lens marker’s equation. The focal length of the lens is positive for a converging lens and negative for a diverging lens. The converging lens can form a virtual image as well as a real one. If the object is outside the focal point, then it is a real image, and if the object is inside the focal point, then it is a virtual image.

Formulae:

The lens formula for refraction,

1f=n-11r1-1r2 (i)

The lens formula, 1f=1P+1i(ii)

The magnificance formula of the lens, m=-iP(iii)

Here f is the focal length, iis the image distance, p is the object distance, m is the magnification.

03

(a) Calculation of the object distance

In the given problem, r1is positive and r2is negative, hence the given lens is of double-convex type.

Using the given data in equation (i), the expression of the focal length of the lens in air is given as follows:

f=r1r2n-1r2-r1=+30cm-30cm1.50-1-30cm-+30cm=+30cm

The given lens is a converging lens because its focal length is positive.

For an object in front of the lens, the object distance Pand image distance i are related to the lens’ focal length. Thus, the image distance can be given using the data in equation (ii) as follows:

1i=1f-1Pi=PfP-f=+10cm+30cm+10cm-+30cm=-15cm

Hence, the value of the image distance is-15cm .

04

(b) Calculation of the lateral magnification of the object

The lateral magnification is the ratio of the object distance Pto the image distance i. It is given using the data in equation (iii) as follows:

m=--15cm+10cm=+1.5

Hence, the value of the lateral magnification is +1.5.

05

(c) Calculation of the behavior of the image

If the object is inside the focal point, then it is a virtual image.The image distance is also negative.

Hence, the image is virtual.

06

(d) Calculation if the image is inverted or not

The value of magnification is positive.

Hence, the image is not inverted .

07

(e) Calculation of the position of the image

The image distance is negative.

Hence, the image is on the same side of the object.

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Most popular questions from this chapter

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?

An object is placed against the center of a converging lens and then moved along the central axis until it is 5.0mfrom the lens. During the motion, the distance between the lens and the image it produces is measured. The procedure is then repeated with a diverging lens. Which of the curves in Fig. 34-28 best gives versus the object distance p for these lenses? (Curve 1 consists of two segments. Curve 3 is straight.)

Figure 34-50a is an overhead view of two vertical plane mirrors with an object O placed between them. If you look into the mirrors, you see multiple images of O. You can find them by drawing the reflection in each mirror of the angular region between the mirrors, as is done in Fig. 34-50b for the left-hand mirror. Then draw the reflection of the reflection. Continue this on the left and on the right until the reflections meet or overlap at the rear of the mirrors. Then you can count the number of images of O. How many images of O would you see if θis (a) 90°, (b) 45°, and (c) 60°? If θ=120°, determine the (d) smallest and (e) largest number of images that can be seen, depending on your perspective and the location of O. (f) In each situation, draw the image locations and orientations as in Fig. 34-50b.

A moth at about eye level is10cmin front of a plane mirror; a man is behind the moth,30cmfrom the mirror. What is the distance between man’s eyes and the apparent position of the moth’s image in the mirror?

A fruit fly of height H sits in front of lens 1 on the central axis through the lens. The lens forms an image of the fly at a distance d=20cmfrom the fly; the image has the fly’s orientation and height H1=2.0H. What are (a) the focal lengthf1 of the lens and (b) the object distance p1of the fly? The fly then leaves lens 1 and sits in front of lens 2, which also forms an image at d=20cmthat has the same orientation as the fly, but now H1=0.50H. What are (c) f2and (d) p2?

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