Q 81.

Expert-verifiedFound in: Page 994

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

A fortune teller’s “crystal ball” (actually just glass) is 10 cm in diameter. Her secret ring is placed 6.0 cm from the edge of the ball.

a. An image of the ring appears on the opposite side of the crystal ball. How far is the image from the center of the ball?

b. Draw a ray diagram showing the formation of the image.

c. The crystal ball is removed and a thin lens is placed where the center of the ball had been. If the image is still in the same position, what is the focal length of the lens?

(a) The distance between the image and the center of the ball is 24cm.

(b) The diagram is made by using image formation principles.

(c) f=7.5cm

An image can be formed by refraction from a spherical surface of radius RR. The object s and image s' distances from such a surface are related by

where the light is incident in the medium for which the index of refraction is and is refracted in the medium for which the index of refraction is . The radius of curvature R is positive for surfaces convex toward the object, negative otherwise.

2- The distance s of an object from a lens, the distance s′ of the image from the lens, and the focal length ff of the lens are related by the thin lens equation:

Several sign conventions are important when using the thin lens equation:

The focal length f is positive for convex lenses and negative for concave lenses.

The image distance s′ is positive for real images and negative for virtual images.

The diameter of the ball is: D=10cm.

The radius of the ball is then: R=5.0cm.

The distance between the ring and the edge of the ball is: s=6.0cm.

The index of refraction of glass is: n=1.50.

The index of refraction of the air is: n=1.00.

Determine the distance between the image of the crystal ball and the center of the ball.

The crystal ball can be considered as a combination of two equally curved surfaces each with a radius of curvature R, where . Assuming the ring is located on the left, then the first surface acts as a convex surface with a positive radius of curvature and the second surface acts as a concave surface with a negative radius of curvature

First, we consider the image formed by surface 1. Using Equation (1), we find that image 1 formed by surface 1 satisfies the following equation

where =1.00 is the index of refraction of the surrounding medium which is air, =6.0cm is the distance between the object and surface 1, =1.50 is the index of refraction of the material of the crystal ball, is the distance between image 1 and surface 1, and =+5.0cm is the radius of curvature of surface 1. Substituting these values into the above expression, we obtain:

Rearrange and solve for

where the negative sign indicates that image 1 is negative.

Then, we consider the image formed by surface 2. Image 1 now acts as the object. Using Equation (1), we find that image 2 formed by surface 2 satisfies the following equation

Notice the switch in and because the light rays approaching surface 2 are in the material of the lens, and this material has index of refraction

Figure 1 shows that , measured from surface 2, is related to as , where is the diameter of the ball. Because is negative and is positive. Hence, the previous equation becomes

Substitute numerical values:

Rearrange and solve for

From Figure 1, we see that the distance between the image and the center of the ball is: 19 cm +5.0 cm=24 cm

Draw a ray diagram showing the formation of the image.

The below diagram shows the formation of the image.

Determine the focal length of a thin lens that is put in the location of the center of the ball if the image is still in the same position.

If the crystal ball is replaced by a thin lens, the object distance would be:

, and the image distance would be

The focal length of the lens is then found by using the thin lens equation ``Equation (2)":

Substitute numerical values:

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