A microscope has an objective lens of 10mm focal lenth and eye piece of 25.0mm focal length. What is the distance between the lenses and its magnification if the object is in Sharp focus when it is 10.5mm from the objective
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