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Questions: 1 2 3 4 5 6 7 8 9 10 11

Optical Instruments Solutions

  1. The nearpoint of a farsighted person is 125 cm. What power contact lenses must this person wear to have "normal" vision?

    For this farsighted person, an object must be 125 cm away from his eyes if he is to see it clearly. For a "normal" person, the nearpoint is assumed to be 25 cm. The contact lens must cause an image of the object at 25 cm appear to be 125 cm on the incident side. This means that the image is erect and virtual.

    We have o = 25 cm and i = -125 cm. Using the lens equation

    1/f = 1/o + 1/i = 1/25 - 1/125 = 4/125 .

    Thus the focal length of the lens must be f = +31.25 cm. The lens is positive or converging.

    The power of a lens is defined

    P = 1/f = 1/0.3125 m = 3.20 diopters .

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  2. If the person in Question #1 prefers glasses, what power should they be? Assume that the glasses are 2.00 cm from the eyes.

    Accounting for the position of the glasses, the object would be o = 25 cm - 2 cm = 23 cm and i = -(125 cm - 2 cm) = -123 cm.

    Using the lens equation

    1/f = 1/o + 1/i = 1/23 - 1/123 = 0.03535 cm .

    Thus the focal length of the lens must be f = +28.29 cm. The lens is positive or converging.

    The power of a lens is defined

    P = 1/f = 1/0.2829 m = 3.53 diopters .

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  3. If the person in Question #1 tries on contact lenses of power +2.5 diopters, where would his corrected nearpoint be?

    We are asking how close an object can be in front of the person’s eye, o, if the final image is at 125 cm, or i = –125 cm. The minus sign is required since the image is on the incident side of the eye. We are given P = 1/f = 2.5 m–1, so f = 40 cm. Solving for o,

    1/o = 1/f – 1/i = 1/40 – 1/(–125).

    This yields o = 30.30 cm.

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  4. The farpoint of a nearsighted person is 8.00 cm. What minimum power contact lenses should the person wear to be able to read a newspaper held 50 cm away from the eye?

    For this nearsighted person, an object can be no farther than 8 cm away from his eyes if he is to see it clearly. If the object, the newpaper, is 50 cm away, the person needs the image to be 8 cm away on the incident or object side of the lens. This means that the image is erect and virtual.

    We have o = 50 cm and i = -8 cm. Using the lens equation

    1/f = 1/o + 1/i = 1/50 - 1/8 = -21/200 .

    Thus the focal length of the lens must be f = -9.52 cm. The lens is negative or diverging.

    The power of a lens is defined

    P = 1/f = -1/0.09524 m = -10.5 diopters .

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  5. If the person in Question #4 prefers glasses, what power should they be. Assume that the glasses are 2.00 cm from the eyes.

    Accounting for the position of the glasses, the object would be o = 50 cm - 2 cm = 48 cm and i = -(8 cm - 2 cm) = -6 cm.

    Using the lens equation

    1/f = 1/o + 1/i = 1/48 - 1/6 = -7/48 .

    Thus the focal length of the lens must be f = -6.86 cm. The lens is negative or diverging.

    The power of a lens is defined

    P = 1/f = -1/0.06857 m = -14.6 diopters .

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  6. A nearsighted person has a farpoint of 30.0 cm and a nearpoint of 8.5 cm. (a) What power should his contact lenses be? (b) What will be his corrected farpoint and nearpoint?

    The contact lenses correct his nearsightedness, that is it allows him to see distant objects, o = ∞.  The lenses must form images just inside his original farpoint. Since the image is on the incident sides, i = –30.0 cm. Solving for the focal length and then the power

    1/f = 1/o + 1/i = 1/∞ + 1/(–30) = 1/(–30)

    the focal length is f = –30 cm and the power P = 1/f = –3.33 m-1.

    Wearing the lenses affects how close a person can hold an object to his eye and see it clearly, that is his nearpoint is changed. We are trying to find the object distance o and the final image must be at the person’s uncorrected nearpoint. Since that image is on the incident side i = –8.5. Solving for o,

    1/o = 1/f  – 1/i = 1/(–30) – 1/(–8.5) = 1/11.86 .

    The “corrected” nearpoint is 11.86 cm.

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  7. The nearsighted person in the previous problem puts in contacts of power –2.25 Diopters, what is his corrected farpoint and nearpoint?

    To find the corrected farpoint, we find out how far away an object can be with this lens and still have an image at the uncorrected farpoint. The image is on the incident side so the sign convention requires the addition of a minus sign, i.e. i = –30.0 cm. The focal length of the length is f = 1/P = 1/(–2.25 m-1) = –44.44 cm. Solving for o,

    1/o = 1/f  – 1/i = 1/(–44.44) – 1/9–30.0)

    which yields o = 92.31 cm. The corrected farpoint is 92.31 cm.

    To find the corrected nearpoint, we find out how far away an object can be with this lens and still have an image at the uncorrected nearpoint. The image is on the incident side so the sign convention requires the addition of a minus sign, i.e. i = –8.5 cm. The focal length of the lens is f = 1/P = 1/(–2.25 m-1) = –44.44 cm. Solving for o,

    1/o = 1/f  – 1/i = 1/(–44.44) – 1/(–8.5)

    which yields o = 10.51 cm. The corrected nearpoint is 10.51 cm.

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  8. A nearsighted person wears contacts of –3.50 diopters and can clearly see objects held 14.0 cm away. What is the uncorrected farpoint and nearpoint of this person?

    The focal length of the lens is f = 1/P = 1/(–3.50 m-1) = –28.57 cm.

    These lenses correct the persons nearsightedness meaning that with the lenses she can see objects at o = ∞ and the resulting images will be at the persons uncorrected farpoint so that i = –XFP. The minus sign indicates that the image is on the incident side of the lens. Solving for XFP

    1/i = 1/–XFP = 1/f – 1/o = 1/–28.57 – 1/∞

     which yields XFP = 28.57 cm.

    The person’s corrected nearpoint is indicated to be 14.0 cm. This means an object at this point, o = 14.0 cm, must have an image at the uncorrected nearpoint, i = –XNP. The minus sign indicates that the image is on the incident side of the lens. Solving for XNP

    1/i = 1/–XNP = 1/f – 1/o = 1/(–28.57) – 1/14

    which yields XNP = 9.40 cm.

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  9. A photographer is taking pictures on a bright sunny day with a camera shutter speed of 1/250 s and f/5.6. Suddenly the sky clouds over and the available light is reduced by three-quarters. (a) If only the shutter speed is changed, what new shutter speed would be needed for proper contrast? (b) If only the f-number is changed, what new f-number is needed?

    Let I0 be the initial light intensity, then the later intensity when it is cloudy is I = ¼I0.

    To properly expose film requires that a specific amount of light reaches the film. The amount of light reaching the film is governed by

    Lfilm µ ID2t ,

    where I is the intensity of the ambient light, D2 is a measure of the area of the aperture, and t is the time that the aperture. The area is set by the f-number or f-stop of the camera. The time is set by the shutter speed. Our adjustments of the camera must be such that Lfilm is a constant.

    (a) Changing only the shutter speed:

    Linitial = Lfinal
    I0D2t0 = (¼I0)D2tf

    Solving for tf, we find

    tf = 4t0 = 4(1/250 s) = 1/62.5 s ≈ 1/60 s .

    The nearest available shutter setting would actually be 1/60th s.

    (b) Changing only the f-number:

    Linitial = Lfinal
    I0(D0)2t0 = (¼I0)(Df)2t0

    Solving for Df, we find

    (Df)2 = 4(D0)2 .

    So we need to increase the aperture area by a factor of 4. Hence we must move two f-stops, since the area changes by a factor of two with each change of f-stop. As the f-stop number gets smaller the area gets bigger, so we need f/2.8.

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  10. The separation of the lenses in a compound microscope is 18.0 cm. The objective and eyepiece have focal lengths of 0.20 cm and 3.00 cm, respectively. (a) What magnification would a person with normal vision observe? (b) Where would the object have to be placed?

    Let d = 18.0 cm.

    (a) The magnification of a telescope is given by

    M = -(L/f0)(XNP/fe) ,

    where L = d - f0 - fe = 18.0 cm - 3.00 cm - 0.20 cm = 14.8 cm and XNP is the nearpoint of the "normal" person which is set at 25 cm. Thus the magnification is

    M = -(14.8/0.2)(25/3) = -617 ,

    The minus sign indicated that the image will be inverted.

    (b) In a telescope the object is placed in front of the objective so that the image created by the objective is formed at the focal point of the eyepiece.

    We have fo = 0.20 cm and we want i = d - fe = 18.0 cm - 3.00 cm = 15.0 cm. Note that since the image is on the transmission side of the objective it is real and I is positive. Using the lens equation,

    1/o = 1/f - 1/i = 1/0.20 - 1/15 = +4.933 .

    Thus the object must be placed 0.203 cm in front of the objective.

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  11. Galileo's telescope had a magnification of 3. Assuming that the focal length of the eyepiece was 15 cm, how long was his telescope?

    The length of the telescope is given by L = f0 + fe.

    We are told that

    M = -f0/fe = -3 .

    Thus fo = 3fe, and

    L = 4fe = 4(15 cm) = 60 cm.

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