23.18. Model: Represent the Can As a Point Source and Use the Ray Model of Light

Physics III

Homework V CJ

Chapter 23; 18, 24, 32, 54, 60, 65, 70, 78

23.18. Model: Represent the can as a point source and use the ray model of light.

Visualize: /

Paraxial rays from the can refract into the water and enter into the fish’s eye.

Solve: The object distance from the edge of the aquarium is s. From the water side, the can appears to be at an image distance s 30 cm. Using Equation 23.13,

23.24. Model: Use the ray model of light and the phenomenon of dispersion.

Visualize: /

Solve: Using Snell’s law for the red light,

Now using Snell’s law for the violet light,

nviolet 1.48

Assess: As expected, nviolet is slightly larger than nred.

23.32. Model: Assume the biconcave lens is a thin lens.

Visualize: Please refer to Figure Ex23.32.

Solve: If the object is on the left, then the first surface has R140 cm (concave toward the object) and the second surface has R240 cm (convex toward the object). The index of refraction of glass is 1.50, so the lens-maker’s equation is

f40 cm

23.54. Model: Use the ray model of light and the phenomena of refraction and dispersion.

Visualize: /

Solve: Since violet light is perpendicular to the second surface, it must reflect at violet 30 at the first surface. Using Snell’s law at the air-glass boundary where the ray is incident,

Since nviolet 1.02 nred, . Using Snell’s law for the red light at the first surface

The angle of incidence on the rear face of the prism is thus r glass 30.664 30 0.664. Using Snell’s law once again for the rear face and for the red wavelength,

Becausev air 0 and r air 0.997, .

23.60. Model: Use ray tracing to locate the image. Assume that the converging lens is a thin lens.

Solve: (a) /

The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three special rays that experience refraction do not converge at a point. Instead they appear to come from a point that is 15 cm on the same side as the object itself. Thus s15 cm. The image is upright and has a height of h 1.5 cm.

(b) Using the thin-lens formula,

The image height is obtained from

The image is upright and 1.5 times the object, that is, 1.5 cm high. These values agree with those obtained in part (a).

23.65. Model: Assume the lens is a thin lens and the thin-lens formula applies.

Solve: Because we want to form an image of the spider on the wall, the image is real and we need a converging lens. That is, both s and s are positive. This also implies that the spider’s image is inverted, so . Using the thin-lens formula with

We also know that the spider is 2.0 m from the wall, so

ss 2.0 m s

Thus, and s 2.0 m –1.33 m 0.67 m 6.7cm. We need a 44.4 cm focal length lens placed 67cm from the wall.

23.70. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens.

Visualize:

The figure shows the two lenses and a ray-tracing diagram. The ray-tracing shows that the lens combination will produce a real, inverted image behind the second lens.

Solve: (a) From the ray-tracing diagram, we find that the image is 50 cm from the second lens and the height of the final image is 4.5 cm.

(b)s1 15 cm is the object distance of the first lens. Its image, which is a virtual image, is found from the thin-lens equation:

The magnification of the first lens is

The image of the first lens is now the object for the second lens. The object distance is s2 24 cm  10 cm  34 cm. A second application of the thin-lens equation yields:

The magnification of the second lens is

The combined magnification is . The height of the final image is (2.286)(2.0cm)  4.57 cm. These calculated values are in agreement with those found in part (a).

23.78. Model: Assume the ray model of light. The ball is not a thin lens. However, the image due to refraction from the first surface is the object for the second surface.

Visualize: /

Solve: (a) For refraction from the first surface, R5 cm (convex toward the object). Thus,

The image is virtual (to the left of the surface) and upright.

For refraction from the second surface, s2 22.5 cm  10 cm  32.5 cm and R5.0 cm (concave toward the object). Thus,

The image is 18.6 cm from the right edge of the ball and thus 23.6 cm from the center.

(b) The ray diagram showing the formation of the image is shown above.

f 7.5 cm