The pocket microscope used by a student consists of eye lens of focal length 6.25cm and objective of focal length 2cm. At microscope length 15cm, the final image appears biggest. Estimate distance of the object from the objective and magnifying power of the microscope.
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
object 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
Explanation:
hope it helps you