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Yakvenalex [24]
3 years ago
11

A typical person's eye is 2.5 cm in diameter and has a near point (the closest an object can be and still be seen in focus) of 2

5 cm, and a far point (the farthest an object can be and still be in focus) of infinity. 1) What is the minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye? (Express your answer to two significant figures.)
Physics
1 answer:
Scrat [10]3 years ago
6 0

Answer:

The minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye is 2.27 cm.

Explanation:

The diameter of person's eye is 2.5 cm. The close point or the near point of the eye is 25 cm and the far point is infinity. We need to determine the minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye. Let f is the minimum effective focal length. It can be calculated using lens formula as :

\dfrac{1}{f}=\dfrac{1}{25}+\dfrac{1}{2.5}

f = 2.27 cm

So, the minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye is 2.27 cm and it is at the nearest point. Hence, this is the required solution.

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Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sp
Leokris [45]

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ q_{i} / r_{i}

where q_{i} and r_{i} are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

6 0
3 years ago
relationship beetween image distance,focal length ,power magnification, for both concave and convex lens.​
scoray [572]

Answer:

Object distance means what is the distance between pole and object. Image distance means when image is formed then the distance between pole and image is called image distance. Focal length is the distance between pole and the principal focus of the mirror.

A lens is a clear object, usually made of glass or plastic, which is used to refract, or bend light. Lenses can concentrate light rays (bring them together) or spread them out. Common examples of lenses include camera lenses, telescope lenses, eyeglasses, and magnifying glasses. Lenses are often double lenses, meaning they have two curved sides. A convex lens is rounded outward, while a concave lens curves inward. (A great way to remember this is that a concave lens creates an indent like a cave!)

The image distance can be calculated with the knowledge of object distance and focal length with the help of lens formula. In optics, the relationship between the distance of an image (i), the distance of an object (o), and the focal length (f) of the lens are given by the formula known as Lens formula. Lens formula is applicable for convex as well as concave lenses. These lenses have negligible thickness. It is an equation that relates the focal length, image distance, and object distance for a spherical mirror. It is given as,

1/i + 1/o = 1/f

i= distance of the image from the lens

o= distance of the object from the lens

f= focal length of the lens

Explanation:

Hope it is helpful....

7 0
3 years ago
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bonufazy [111]
Since,
Speed = Frequency * WaveLength

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Plug in the values in equation(A):

A => WaveLength = \frac{3 * 10^{8}}{13*10^{3}}

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Answer:

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Explanation:

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Bingel [31]
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NOTE that in Physical Change there is no change in the chemical structure and the material retains all its chemical properties, and no new compounds are produced. 

again, A physical change is any change not involving a change in the substance's chemical identity. Matter undergoes chemical change when the composition of the substances changes: one or more substances combine or break up (as in a relationship) to form new substances.Physical changes occur when objects undergo a change that does not change their chemical nature. A physical change involves a change in physical properties. Physical properties can be observed without changing the type of matter. Examples of physical properties include: texture, shape, size, color, odor, volume, mass, weight, and density. 

BUT in Chemical Change ( or Chemical Reaction ) there would be change in the chemical nature of the material undergoing a Chemical Change with the production of new compounds.
5 0
3 years ago
Read 2 more answers
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