An object is placed at a distance 1cm from objective of compound microscope.Distance between lenses is 30cm and the intermediate image at is 5cm distance from eyepiece find magnifying power.
Answers
Answer:
Focal length of the objective lens f
1
=2cm
Focal length of the eyepiece,f
2
=6.25cm
Distance between the objective lens and the eyepiece,d=15cm
Least distance of distinct vision,d
′
=25cm
Image distance for the eyepiece,v
2
=−25cm
bject distance for the eyepiece = u
2
According to the lens formula
f
2
1
=
v
2
1
−
u
2
1
6.25
1
=
−25
1
−
u
2
1
u
2
=−5cm
Image distance for the objective lens v
1
=d+u
2
=15−5=10cm
According to the lens formula
f
1
1
=
v
1
1
−
u
1
1
2
1
=
10
1
−
u
1
1
u
2
=−2.5cm
The magnifying power of a compound microscope is given by the relation
m=
∣u
1
∣
v
1
(1+
f
2
d
′
)
m=
2.5
10
(1+
6.25
25
)=20
b)The final image is formed at infinity
Image distance for the eyepiece v
2
=∞
According to the lens formula
f
2
1
=
v
2
1
−
u
2
1
According to the lens formula
6.25
1
=
∞
1
−
u
2
1
u
2
=−6.25cm
Image distance for the objective lens v
1
=d+u
2
=10−6.25=8.75
Object distance for the objective len =u
2
According to the lens formula
f
1
1
=
v
1
1
−
u
1
1
2
1
=
8.75
1
−
u
1
1
u
2
=−2.59cm
The magnifying power of a compound microscope is given by the relation:
∣u
1
∣
v
1
(
∣u
2
∣
d
′
)=13.51