Homework #2: Lenses




Дата канвертавання22.04.2016
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Homework #2: Lenses






1. A contact lens is made of plastic with an index of refraction of 1.50. The lens has an outer radius of curvature of +2.00 cm and an inner radius of curvature of +2.50 cm. What is the focal length of the lens?

2. The left face of a biconvex lens has a radius of curvature of 12.0 cm, and the right face has a radius of curvature of 18.0 cm. The index of refraction of the glass is 1.44. (a) Calculate the focal length of the lens. (b) Calculate the focal length if the radii of curvature of the two faces are interchanged.


3. A converging lens has a focal length of 20.0 cm. Locate the images for object distances of (a) 40.0 cm, (b) 20.0 cm, and (c) 10.0 cm. For each case, state whether the image is real or virtual and upright or inverted, and find the magnification.


4. A diverging lens has a focal length of 20.0 cm. Locate the images for object distances of (a) 40.0 cm, (b) 20.0 cm, and (c) 10.0 cm. For each case, state whether the image is real or virtual and upright or inverted, and find the magnification.

5. A transparent photographic slide is placed in front of a converging lens with a focal length of 2.44 cm. The lens forms an image of the slide 12.9 cm from it. How far is the lens from the slide if the image is (a) real? (b) virtual?

6. A certain LCD projector contains a single thin lens. An object 24.0 mm high is to be projected so that its image fills a screen 1.80 m high. The object-to-screen distance is 3.00 m. (a) Determine the focal length of the projection lens. (b) How far from the object should the lens of the projector be placed in order to form the image on the screen? {Hint: How are p and q related to the 1.80 meters? Try p + q = 1.80 m. Combine this relationship and the definition of magnification to solve part (a).}


7. A person uses a converging lens that has a focal length of 12.5 cm to inspect a gem. The lens forms a virtual image 30.0 cm away. Determine the magnification. Is the image upright or inverted?


8. An object is 5.00 m to the left of a flat screen. A converging lens for which the focal length is f = 0.800 m is placed between object and screen. (a) Show that there are two lens positions that form an image on the screen, and determine how far these positions are from the object. (b) How do the two images differ from each other? {Hint: How are p and q related to the 5.00 meters?}


9. {BONUS type question.} An object is placed 12 cm to the left of a diverging lens of focal length −6.0 cm. A converging lens of focal length 12 cm is placed a distance d to the right of the diverging lens. Find the distance d that places the final image at infinity. {Hint: Let the image of the first lens be the object for the second lens.}



Solutions:

1. With , the lens maker’s equation gives the focal length as



or

2. The lens maker’s equation is used to compute the focal
length in each case.

(a)



(b)

3. From the thin lens equation, , the image distance is found to be

(a) If , then and

The image is

(b) If ,

(c) When , and

The image is

4. From the thin lens equation, , the image distance is found to be

(a) If , then and

The image is

(b) If , then and



The image is

(c) When , and

The image is

5. (a) The real image case is shown in the ray
diagram. Notice that , or
. The thin lens equation, with
, then gives

or

Using the quadratic formula to solve gives

Both are valid solutions for the real image case.



(b) The virtual image case is shown in the second
diagram. Note that in this case, ,
so the thin lens equation gives

or

The quadratic formula then gives

Since the object is real, the negative solution must be rejected leaving .

6. It is desired to form a magnified, real image on the screen using a single thin lens. To do this, a converging lens must be used and the image will be inverted. The magnification then gives

, or

Also, we know that . Therefore, giving

(b)

(a) The thin lens equation then gives

or

7. We are given that . Then, the thin lens equation, , gives



and the lateral magnification is

Since , the image is

8. (a) The total distance from the object to the real image is the object-to-screen distance, so or . The thin lens equation then becomes



or

With , this gives which factors as with two solutions:

(b) If , then and

In this case, the image is

If , then and . In this case, the image is

9. Applying the thin lens equation to the first lens gives

The virtual image formed by the first lens serves as the object for the second lens. If the image formed by that converging lens is to be located at infinity, the image formed by the diverging lens must be located at the focal point in front of the converging lens.



Thus, or


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