9.11: A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Focal length of the objective lens, f1=2.0 cm

The focal length of the eyepiece, f2=6.25 cm

(a)
Hint: Use the lens formula.

Step 1: Find the image distance for the objective lens.
Least distance of distinct vision, d’ = 25 cm

 Image distance for the eyepiece, v2= -25cm

According to the lens formula,

1v2-1u2=1f2
1u2=1v2-1f2
     =1-25-16.25=-1-425=-525
 u2=-5 cm

Image distance for the objective lens, 

Step 2: Find the magnitude of object distance.
According to the lens formula,

1v1-1u1=1f11u1=1v1-1f1      =110-12=1-510=-410 u1=-2.5 cm

Magnitude of the object distance, u1=2.5 cm

Step 3: Find the magnifying power.
The magnifying power of a compound microscope:

m=v1|u1|1+d'f2
   =102.51+256.25=41+4=20

Hence, the magnifying power of the microscope is 20.

(b)
Hint: Use the lens formula.

Step 1: Find the image distance for the objective lens.
When the final image is formed at infinity:

 Image distance for the eyepiece, 

According to the lens formula,

Image distance for the objective lens, 

Step 2: Find the magnitude of object distance.
According to the lens formula,

The magnitude of the object distance, =2.59 cm

Step 3: Find the magnifying power.
The magnifying power of a compound microscope:

m=v1u1d'u2=8.752.59×256.25=13.51

Hence,  the magnifying power of the microscope is 13.51.