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KIM [24]
3 years ago
10

A physicist's right eye is presbyopic (i.e., farsighted). This eye can see clearly only beyond a distance of 97 cm, which makes

it difficult for the physicist to read books and journals. Find the focal length and power of a lens that will correct this presbyopia for a reading distance of 25 cm, when worn 2 cm in front of the eye
Physics
1 answer:
Ivan3 years ago
5 0

Answer:

f = 19,877 cm   and  P = 5D

Explanation:

This is a lens focal length exercise, which must be solved with the optical constructor equation

        1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image.

In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye

let's calculate

        1 / f = 1/97 + 1/25

        1 / f = 0.05

        f = 19,877 cm

the power of a lens is defined by the inverse of the focal length in meters

         P = 1 / f

         P = 1 / 19,877 10-2

         P = 5D

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Explanation:

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- Resonance structure of a molecule is of two or more forms in which the distribution of electrons around the structure is different but the chemical connectivity is same.

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- The formal charge FC on each atom is determined as:

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Where,

            NBE: Non-Bonding Electrons

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- The formal charge on each atom is: the nitrogen atom in center will possess 1+ formal charge and oxygen will possess 1- charge (oxygen is electronegative atom). Thus, results in formation of neutral molecule.

- The structure of (N N O) is shown in attachment.

- The resonance form which is likely to contribute most to the correct structure of (N_2 O) is:

- Structure for (N N O)  showing one lone-pair of electrons on the terminal nitrogen atom, a triple bond between the two nitrogen atoms, a single bond between nitrogen and oxygen, and three lone-pairs of electrons on the terminal oxygen atom.

               

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Answer:

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Explanation:

Dont report my answer please

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