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Chapter 34: Q121P (page 1045)

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.

Short Answer

Expert verified
  1. The image distance i is -12cm.
  2. The ray diagram is drawn showing the image position.

Step by step solution

01

Listing the given quantities

Object distance p=20cm

Focal length of the lens f = -30cm

02

Understanding the concepts of lens equation

Here, we need to use the equation of a lens to calculate the image distance for a diverging lens.

We can draw a ray diagram showing the image position using two rays, the ray passing through the center of lens and other ray passing parallel to the principle axis.

Formula:

1p+1i=1f

03

Calculations of the image distance

(a)

The lens equation relates an object distance p, lens focal length f and the image distance i as

1p+1i=1f1i=1f-1p1i=p-ffpi=pfp-f=(20cm)(-30cm)(20cm--30cm)=-12cm

The negative sign shows that the image isvirtualand to the left side of the lens.

The image distance i is -12cm.

04

Step 4: The ray diagram showing the image position

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Most popular questions from this chapter

32 through 38 37, 38 33, 35 Spherical refracting surfaces. An object Ostands on the central axis of a spherical refracting surface. For this situation, each problem in Table 34-5 refers to the index of refraction n1where 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 the object Oor on the opposite side.

Light travels from point A to B point via reflection at point O on the surface of a mirror. Without using calculus, show that length AOB is a minimum when the angle of incidence 胃is equal to the angle of reflection 蠒.

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 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?


Isaac Newton, having convinced himself (erroneously as it turned out) that chromatic aberration is an inherent property of refracting telescopes, invented the reflecting telescope, shown schematically in Fig. 34-59. He presented his second model of this telescope, with a magnifying power of 38, to the Royal Society (of London), which still has it. In Fig. 34-59, incident light falls, closely parallel to the telescope axis, on the objective mirror. After reflection from the small mirror (the figure is not to scale), the rays form a real, inverted image in the focal plane (the plane perpendicular to the line of sight, at focal point F). This image is then viewed through an eyepiece. (a) Show that the angular magnification for the device is given by Eq. 34-15:

m胃=fob/fey

fob

the focal length of the objective is a mirror and

feyis that of the eyepiece.

(b) The 200 in. mirror in the reflecting telescope at Mt. Palomar in California has a focal length of 16.8 m. Estimate the size of the image formed by this mirror when the object is a meter stick 2.0 km away. Assume parallel incident rays. (c) The mirror of a different reflecting astronomical telescope has an effective radius of curvature of 10 m (鈥渆ffective鈥� because such mirrors are ground to a parabolic rather than a spherical shape, to eliminate spherical aberration defects). To give an angular magnification of 200, what must be the focal length of the eyepiece?
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