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12345 [234]
4 years ago
5

Need help

Physics
1 answer:
Agata [3.3K]4 years ago
5 0

Answer:

2.6 then 1.5 last 0.8

Explanation:

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A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them
Tatiana [17]

Answer:

A)    q₂ = 75.98 cm, B)     q₂' = 115.38 cm, C)

Explanation:

A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.

Lens 1. More to the left

let's use the constructor equation

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

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

We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.

          \frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}

         \frac{1}{q_1} = \frac{1}{14.8} - \frac{1}{50}  

          1 / q₁ = 0.04756

           q₁ = 21.0227 cm

this image is the object for the second lens that has f₂ = 14.8 cm

the distance must be measured from the second lens

          p₂ = 39.4 -q₁

          p₂ = 39.4 -21.0227

          p₂ = 18.38 cm

let's use the constructor equation

            1 / q₂ = 1 / f - 1 / p2

             

             \frac{1}{q_2} = \frac{1}{14.8} - \frac{1}{18.38}

            \frac{1}{q_2} = 0.01316

            q₂ = 75.98 cm

measured from the second lens

B) the position of the final image with respect to the first lens is

            q₂’= q₂ + 39.4

             q₂'= 75.98 +39.4

              q₂' = 115.38 cm

C) the magnification of a lens is

              m = - q / p

in this case the image measured from lens 2 is q2 = 75.98 cm

the distance to the object from the first lens is p1 = 50cm

          m = - 75.98 / 50

          m = -1.5 X

the negative sign indicates that the image is inverted

4 0
3 years ago
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