A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them, and they a

Question

A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them, and they appear concentric when viewed from either end). The lenses are identical, each with a positive (converging) focal length of 14.8 cm. They are separated by a distance of 39.4 cm. Lens 1 is to the left of Lens 2.

Required:
a. What is the final image’s distance (in cm) from Lens 2?
b. Where is the final image located?
c. What is the overall magnification of the lens pair, considered as a single optical instrument?

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Gia Bảo 4 years 2021-08-09T00:53:29+00:00 1 Answers 110 views 0

Answers ( )

  1. niczorrrr
    0
    2021-08-09T00:54:41+00:00
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    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

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