30. In a compound microscope, if P, is the power of objective and P, is the power of eye piece, which of the following is correct a) Pi P2 c) P₂ = P2 d) PSP2
Answers
Answer:
Focal length of the objective lens, f
o
=1.25cm
Focal length of the eyepiece, f
e
=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-
m
e
=(1+d/f
e
)=6
The angular magnification of the objective lens (m
o
) is calculated as-
m
o
m
e
=m⇒m
o
=5
Now, m
o
=−v
o
/u
o
⇒v
o
=−5u
o
Using lens formula,
v
o
1
−
u
o
1
=
f
o
1
Using above two equations, u
o
=−1.5cm and v
o
=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 v
e
=−d=−25cm
Using lens formula,
v
e
1
−
u
e
1
=
f
e
1
u
e
=−4.17cm
Separation between the objective lens and the eyepiece, ∣u
e
∣+∣v
o
∣=11.67cm
Therefore, the separation between the objective lens and the eyepiece should be 11.67cm.