Calculate the magnification of a compound light microscope with an objective lens of focal length 0.9 cm and an eyepience with a focal length of 2.5 cm. Separation between the foci of the two lenses is 10 cm and image is viewed at a distance of 20 cm. Magnification=__
Answers
Answered by
1
Answer:
Focal length of the objective lens,f
o
=8mm=0.8cm
Focal length of the eyepiece, f
e
=2.5cm
Object distance for the objective lens,μ
o
=−9mm=−0.9cm
Least distance of distant vision d=25cm
Image distance for the eyepiece,v
e
=−d=−25cm
Object distance for the eyepiece =u
e
Using the lens formula, we can obtain the value of u
e
v
e
1
−
u
e
1
=
f
e
1
−25
1
−
u
e
1
=
2.5
1
So, u
e
=−2.27cm
We can also obtain the value of the image distance for the objective lens v
o
using the lens formula.
v
o
1
−
u
o
1
=
f
o
1
v
o
1
−
−0.9
1
=
0.8
1
v
o
=7.2 m
Distance between the objective lens and the eye piece is ∣u
e
∣+v
o
so separation is 2.27+7.2=9.47cm
The magnifying power of the microscope is calculated as
m=
∣u
o
∣
v
o
(1+d/f
e
)
⇒
0.9
7.2
(1+
2.5
25
)=88
Similar questions