# . A compound microscope consists of an objective lens of focal length

2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm .

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 (25cm) , and

(b) at infinity? What is the magnifying power of the microscope in each case?

Explanation:

• Focal length of the objective lens fo=2cm
• Focal length of the eyepiece lens fe=6.25cm
• Distance between the objective lens and the eyepiece, L=15cm
• Least distance of distinct vision,D=25cm

• Image distance for the eyepiece, ve=-25cm
• Object distance for the eyepiece, is ue
• Applying lens formula for eyepiece lens,

1/ve-1/ue=1/fe

⇒1/-25-1/ue=1/6.25

⇒ue=-5cm

• Image distance from objective lens is v0=L-|ue|=15-5=10cm
• Now applying lens formula for objective lens,

1/vo-1/uo=1/fo

⇒1/10-1/uo=1/2

⇒uo=-2.5cm

Image distance for the eyepiece, ve=-∞

Object distance for the eyepiece, is ue

Applying lens formula for eyepiece lens,

1/ve-1/ue=1/fe

⇒1/-∞-1/ue=1/6.25

⇒ue=-6.25cm