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lidiya [134]
2 years ago
12

Consider 2 converging lenses of focal lengths 5 mm (objective) and 50 mm.(eyepiece) An object 0.1 mm in size is placed a distanc

e of 5.2 mm from the objective.
1. What is the size and location of the image from the objective? What is the linear magnification of this objective?
2. Treat the image from the objective as an object for the eyepiece. If the eyepiece creates an image at infinity, how far apart are the two lenses?
3. What is the angular magnification of the pair of lenses?
Physics
1 answer:
AleksandrR [38]2 years ago
8 0

Answer:

1)   q₁ = 12.987 cm , b)       L = 17.987 cm , c)      m = 179.87

Explanation:

We can solve the geometric optics exercises with the equation of the constructor

         1 / f = 1 / p + 1 / q

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

Let's apply this equation to our case

1) f = 5mm = 0.5 cm

    p₁ = 5.2 mm = 0.52 cm

    h = 0.1 mm = 0.01 cm

    1 / q₁ = 1 / f- 1 / p

    1 / q₁ = 1 / 0.5 - 1 / 0.52 = 2 - 1.923

    1 / q₁ = 0.077

    q₁ = 12.987 cm

2) in this part they tell us that the eyepiece creates an image at infinity, therefore the object that comes from being at the focal length of the eyepiece

            p₂ = 5 cm

The absolute thing that goes through the two lenses is

           L = q₁ + p₂

           L = 12.987 +5

           L = 17.987 cm

3) This lens configuration forms the so-called microscope, whose expression for the magnifications

           m = -L / f_target 25 cm / f_ocular

           m = - 17.987 / 0.5 25 / 5.0

            m = 179.87

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\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

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The minimum uncertainty in the object's velocity is 0.31\times 10^{-4}\ \text{m/s}.

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