Question

a compound microscope has an objective of a focal length of 10 mm and a tube 100 mm long. An image is produced at 250mm from the eyepiece when the object is 12mm from the objective what is the angular magnification​

Answer 1

Answer:

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

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

Explanation:

Hope this helps:)  ジョセフ