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erma4kov [3.2K]
3 years ago
15

A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while tr

aveling. Fortunately, he has his old pair as a spare.
(a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?
Physics
1 answer:
svp [43]3 years ago
7 0

Answer:

a)   p = 95.66 cm, b) p = 93.13 cm

Explanation:

For this problem we use the  constructor equation

         \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distances to the object and the image, respectively

the power of the lens is

         P = 1 / f

         f = 1 / P

         f = 1 / 2.25

         f = 0.4444 m

the distance to the object is

         \frac{1}{p} = \frac{1}{f} -\frac{1}{q}

the distance to the image is

          q = 85 -2

           q = 83 cm

we must have all the magnitudes in the same units

           f = 0.4444 m = 44.44 cm

we calculate

           \frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}

           1 / p = 0.010454

            p = 95.66 cm

b) if they were contact lenses

            q = 85 cm

            \frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}

             1 / p = 0.107375

             p = 93.13 cm

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Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

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