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Tju [1.3M]
3 years ago
11

Carol is farsighted ( presbyopia) and cannot see objects clearly that are closer to her eyes than about meter. She sees objects

clearly with a relaxed eye when they are distant. What is the refractive power of reading glasses that would allow her to read a book 50 cm away with a relaxed eye

Physics
2 answers:
Crank3 years ago
8 0

Answer:

+1.00 diopter

Explanation:

The power of a lens can be described simply as the reciprocal of the focal length of the lens measured in meters.

But f is unknown, hence we look for the focal length with the formula

1/f = 1/u + 1/v

where u is former near point = 100cm

v is the new intended near point = 50cm

1/f = 1/50 - 1/100

1/f = 1/100

f = 100 cm

Hence we get Power (D) = 1/f

where f = focal length of the lens in meter

From the question, the focal length of the lens = 100cm = 1m

Hence D = 1/1

D = +1.00

Hence the refractive power of the reading glasses that would allow Carol to read a book 50cm away from the relaxed eye will be +1.00 diopters.

egoroff_w [7]3 years ago
7 0

Answer:

1.0 dioptres

Explanation:

Farsightedness is an eye defect in which a person can see far objects clearly but not near objects. That implies that the patients' near point is farther than 25cm which is the normal least distance of distinct vision.

Farsightedness results from the eyeball being too long or the crystalline lens not being sufficiently converging.

Carol is farsighted with a near point of about a meter (100cm). We desire to make a lens to enable her near point be reduced to about 50cm. The focal length and power of this lens is calculated in the image attached.

The power of a lens is the inverse of its focal length in meters hence the 100 in the formula for power of the lens.

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2 years ago
Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
Masja [62]

Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

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Lostsunrise [7]

Answer:

Corpuscular theory of light

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zimovet [89]

5m

Explanation:

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  Work done = weight x height of lifting

Height of lifting = \frac{work done }{weight}

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The vertical height through which the object was lifted is 5m

learn more:

Work done brainly.com/question/9100769

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