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# Physics 112 Homework 10 (Ch23 &24) Due July 22

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Physics 112 Homework 10 (Ch23 &24)

Due July 22

1. A sharp image is located 78.0 mm behind a 65.0-mm-focal-length converging lens. Find the object distance (a) using a ray diagram, (b) by calculation.

Solution

(a) From the ray diagram, the object distance is about six focal lengths, or 390 mm.

(b) We find the object distance from equation:

which gives

1. A stamp collector uses a converging lens with focal length 24 cm to view a stamp 18 cm in front of the lens. (a) Where is the image located? (b) What is the magnification?

Solution

1. We locate the image from equation:

which gives

The negative sign means the image is 72 cm behind the lens (virtual).

1. We find the magnification from equation:

1. An object is located 1.5 m from an 8.0-D lens. By how much does the image move if the object is moved (a) 1.0 m closer to the lens, and (b) 1.0 m farther from the lens?

Solution

We find the image distance from equation:

which gives
(a) With the new image distance is determined by

which gives

The image has moved or 3.0 cm away from the lens.

(b) With the new image distance is determined by

which gives

The image has moved or 0.5 cm toward the lens.

A 2.00-cm-high insect is 1.20 m from a 135-mm-focal-length lens. Where is the image, how high is it, and what type is it? (b) What if
Solution

(a) We find the image distance from equation:

which gives

We find the height of the image from equation:

which gives

(b) We find the image distance from equation:

which gives

We find the height of the image from equation:

which gives

1. ## Both surfaces of a double convex lens have radii of 31.0 cm. If the focal length is 28.9 cm, what is the index of refraction of the lens material?

Solution

We find the index from the lensmaker’s equation:

Note! We using different sign convention than the Giancoli book.

1. ## A prescription for a corrective lens calls for The lensmaker grinds the lens from a “blank” with and a preformed convex front surface of radius of curvature of 40.0 cm. What should be the radius of curvature of the other surface?

Solution

We find the radius from the lensmaker’s equation:

;

Note! We using different sign convention than the Giancoli book.

1. A ray of light is refracted through three different materials (see figure below). Rank the materials according to their index of refraction, least to greatest.

Solution

By looking at the direction and the relative amount that the light rays bend at each interface, we can infer the relative sizes of the indices of refraction in the different materials (bends toward normal = faster material to slower material or smaller n material to larger n material; bends away from normal = slower material to faster material or larger n material to smaller n material). From the first material to the second material the ray bends toward the normal, thus it slows down and n1 < n2. From the second material to the third material the ray bends away from the normal, thus it speeds up and n2 > n3. Careful inspection shows that the ray in the third material does not bend back away from the normal as far as the ray was in the first material, thus the speed in the first material is the faster than in the third material and n1 < n3. Thus, the overall ranking of indices of refraction is: n1 < n3 < n2.

1. Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a screen 5.00 m away are 6.5 cm apart near the center of the pattern. Determine the wavelength and frequency of the light.

Solution

We find the location on the screen from equation:

For small angles, we have:

which gives

For adjacent fringes, so we have

; which gives
The frequency is

1. In a double-slit experiment, it is found that blue light of wavelength 460 nm gives a second-order maximum at a certain location on the screen. What wavelength of visible light would have a minimum at the same location?

Solution

For constructive interference of the second order for the blue light, we have

For destructive interference of the other light, we have

.

When the two angles are equal, we get

.

For the first three values of we get

which gives

which gives

which gives

The only one of these that is visible light is

1. A light beam strikes a piece of glass at a 60.00° incident angle. The beam contains two wavelengths, 450.0 nm and 700.0 nm, for which the index of refraction of the glass is 1.4820 and 1.4742, respectively. What is the angle between the two refracted beams?

Solution

We find the angles of refraction in the glass from

which gives

which gives

Thus the angle between the refracted beams is

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