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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 objector on the opposite side.

Short Answer

Expert verified

a) The index of refraction n2on the other side of the refracting surface is

b) The object distance p is +71 cm.

c) The radius of curvature r of the surface is +30 cm.

d) The image distance i is +600 cm.

e) The image is real and inverted

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

Step by step solution

01

Step 1: Given

n1=10n2=1.5r=+3cmi=400cm

02

Determining the concept

The index of refraction of object and image, the image distance, and the radius of curvature are given in the problem. Using this data and equation, find the object distance and check whether the image is real or virtual and find the position of the image.

Formulae are as follows:

n1p+n2i=n2-n1T

where p is the pole and i is the image distance.

03

Determining the index of refraction  on the other side of the refracting surface

(a)

The index of refractionon the other side of the refracting surfaceis given in the table 34-5. So,n2=1.5

Therefore, the index of refraction on the other side of the refracting surface is 1.5.

04

Determining the object distance

b)

From equation 34-8

n1p+n2i=n2-n1r

Rearranging the terms,

role="math" localid="1663048347048" n1p=n2-n1r-n2ip=n1n2-n1r-n2i

Substituting the given values

p=1.01.5-1.0+30-1.5+600p=+70.58≈71cm

Therefore, the object distance p is +71 cm.

05

 Determining the radius of curvature r of the surface

(c)

The radius of curvature is given in the problem, r = +30 cm.

Therefore, the radius of curvature r of the surface is +30 cm.

06

Determining the image distance i

(d)

The image distance is given in the problem, i = +600 cm.

Therefore, the image distance i is +600 cm.

07

Determining whether the image is real or virtual

(e)

Since i < 0, the image is real and inverted.

Therefore, the image is real and inverted.

08

Determining the position of the image

(f)

For spherical refracting surfaces, real images form on the opposite side of the object and virtual images form on the same side as the object.

Since the image is real, therefore theimage is on theoppositeside as that of the object.

Therefore, the image is on the opposite side of the object.

The required quantities can be found by using the relation between the index of refraction of object and image, image distance, object distance, and the radius of curvature.

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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?

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