Question

Q 26:  Determine the lens separation and object location for a microscope made from an objective lens of focal length +0.70-cm and an eyepiece of focal length +4.0-cm. Arrange the lenses so that a final virtual image is formed 100 cm to the left of the eyepiece and so that the angular magnification is -260 for a person with a near point of 25 cm.

Part A:  Determine the object distance from the objective lens.

Part B: Determine the distance between the objective lens and the eyepiece.

Answer 1

The given question can be depicted as above. In the figure, the notations are as given below

s1 = Object Distance

s2 = Distance of the image formed by the objective from the eyepiece

s1' = Distance between the objective and the image formed by the objective

s2' = Distance between the final image and eyepiece = -100cm

(Negative sign means that the image is virtual and located to the left of the lens)

fo = Focal length of the objective = +0.70cm

fe = Focal length of the eyepiece = +4cm

d = distance between the two lens = s1' + s2

M = Angular magnification of the microscope = -260

The thin lens equation for the objective is given by

The thin lens equation for the eyepiece is given by

The angular magnification of the microscope is given by

  

Substituting the given values in equation (2)

​​​​​​

  ​​​​​​​​​​​​​

Substitute the now available values in equation (3)

   ​​​​​​​

   ​​​​​​​​​​​​​​

   ​​​​​​​​​​​​​​​​​​​​​

Equation (1) is given by

Multiplying throughout by s1' will give

  

  ​​​​​​​

s1' = 7.688 cm

Now distance between the two lens, d =  s1' + s2

= 7.688 + 3.846

= 11.534 cm (Ans.)

Now subsitute the values obtained above into equation (1)

   

   Therefore s1 = 0.770 cm (Ans.)