Question

An angular magnification (magnifying power) of $30X$ is desire using an objective of focal length $1.25cm$ and an eye piece focal length $5cm$. How will you setup the compound microscope for normal adjustment?

• A. Separation between both lenses $11.67cm$
• B. Separation between both lenses $15.51cm$
• C. Separation between both lenses $6.25cm$
• D. Separation between both lenses $9.0cm$

Answer 1

Answer: (A)

Focal length of the objective lens, fo=1.25cm

Focal length of the eyepiece, fe=5cm

Least distance of distinct vision, d=25cm

Angular magnification of the compound microscope = 30X

Total magnifying power of the compound microscope, m=30

The angular magnification of the eyepiece is given by the relation- me=(1+d/fe)=6

The angular magnification of the objective lens (mo) is calculated as- mome=m⇒mo=5

Now, mo=−vo/uo⇒vo=−5uo

​Using lens formula,

1/vo-1/uo=1/fo
Using above two equations, uo=−1.5cm and vo=7.5cm

So, The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.

Now, Image distance for the eyepiece is ve=−d=−25cm

Using lens formula, 1/ve-1/ue=1/fe

ue= -4.17 cm

Separation between the objective lens and the eyepiece, ∣ue∣+∣vo∣=11.67cm

Therefore, the separation between the objective lens and the eyepiece should be 11.67cm.
So, the correct option is A.