The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye.
Answers
Given :
Focal length of the objective, fo = 1.0 cm
Focal length of the eyepiece, fe = 6 cm
Image distance from the eyepiece, ve = -24 cm
Separation between the objective and the eyepiece = 9.8 cm to 11.8 cm
The lens formula is given by
1/ve-1/ue=1/fe
1/-24-1/ue=1/16
1/ue=-16-12/4
⇒ 1/ue= -4+12/4
=-5/24
∴ ue=-24/5 cm= -4.8 cm
(a) Separation between the objective and the eyepiece = 9.8 cm
So, for the objective lens, the image distance will be
vo = 9.8 − 4.8 = 5 cm
The lens formula for the objective lens is given by
1/v0-1/u0=1/f0
1/5-1/u0=1/1
⇒-1/u0=1-1/5
=5-1/5
⇒u0=-5/4 cm
=-1.25 cm
Magnifying power of the compound microscope:
m=v0/u0 [1+D/fe ]
=51.2/5[ 1+24/6]
=20
(b) Separation between the objective and the eyepiece = 11.8 cm
We have:
m = 30
So, the required range of the magnifying power is 20–30.
Answer:
ANSWER
20 to 30.
For the given compound microscope
f
0
=1cm, f
e
=6cm, D=24cm
For the eye piece, v
e
=−24cm, f
e
=6cm
Now,
v
e
1
−
u
e
1
=
f
e
1
⇒
u
e
1
−
v
e
1
−
f
e
1
⇒−[
24
1
+
6
1
]=−
24
5
⇒u
e
=−4.8cm.
(a) When the separation between objective and eye piece is 9.8cm, the image distance for the objective lens must be (9.8)−(4.8)=5.0cm
Now,
v
0
1
−
u
0
1
=
f
0
1
⇒
u
0
1
=
v
0
1
−
f
0
1
=
5
1
−
1
1
=−
5
4
⇒u
0
=−
4
5
=−1.25cm
So, the magnifying powr is given by,
m=
u
0
v
0
[1+
f
D
]=
−1.25
−5
[1+
6
24
]=4×5=20.
When the separation is 11.8cm,
v
0
=11.8−4.8=7.0cm, f
0
=1cm
⇒
u
0
1
=
v
0
1
−
f
0
1
=
7
1
−
1
1
=−
7
6
So,
m=−
u
0
v
0
[1+
f
D
]=
−(
6
7
)
−7
[1+
6
24
]=6×5=30
So, the range of magnifying power will be 20 to 30.