Question

Where does an object need to be placed relative to a microscope for its 0.500 cm focal length objective to produce a magnification of –400? Where should the 5.00 cm focal length eyepiece be placed to produce a further fourfold (4.00) magnification?

Answer 1

Answer:

(a) $d_{o}= 0.501cm$ and (b) $d_{o}= 3.75cm$

Explanation:

The linear magnification (M) produced by a lens is giving by:

$M = - \frac{d_{i}}{d_{o}}$ (1)

where $d_{i}$: is the image distance and $d_{o}$: is the object distance

Knowing that the lens formula relates the focal length of an image with the distance of the image and the distance of the object:

$\frac{1}{f}= \frac{1}{d_{i}} + \frac{1}{d_{o}}$ (2)

(a) To calculate the distance of the object, first we have to express the equation (1) in function of the image distance:  

$d_{i} = -M \cdot d_{o}$ (3)

Finally, by the introducing $d_{i}$ from (3) into equation (2) we can determine the distance of the object as follows:

$\frac{1}{f}= \frac{1}{-M \cdot d_{o}} + \frac{1}{d_{o}}$

$d_{o} = \frac {f (1 - M)}{-M}$ (4)  

$d_{o} = \frac {0.500 cm (1 - (-400))}{-(-400)} = 0.501 cm$  

So the object needs to be placed at 0.501 cm from the microscope.

(b) To find the distance of the length eyepiece we will do the same treatment as before, to get to equation (4):

$d_{o} = \frac {f (1 - M)}{-M} = \frac {5.00 cm (1 - 4.00)}{-4.00} = 3.75 cm$                                              

Have a nice day!