15.3 Solutions
735
In BFC,
FC
BC
=
sin F ˆ
BC
sin B ˆ
FC
=
sin(π − φ)
sin 2(φ − θ)
=
sin φ
sin 2(φ − θ)
=
φ
2(φ − θ)
(3)
since the angles are small. Hence the equivalent focal length
F eq = FC =
BCφ
2(φ − θ)
=
μr
2(μ − 1)
(4)
where we have used (2), (3) and (4).
Fig. 15.23
15.14 The intensity at a distance r is given by
I =
W
4πr 2 =
100
4π × 5 2 = 0.318 W/m
2
I =
1
2
ε 0 cE
2
0
∴ E 0 =
2 I
ε 0 c
=
2 × 0.318
8.85 × 10 −12 × 3 × 10 8 = 15.48 V/m
15.15 (a) If the object distance is u then the distance v 1 at which the image is
formed by objective is given by
1
u
+
1
v 1
=
1
f o
or v 1 = −
u f o
u − f o
(1)
The distance of this image from the eyepiece, u 1 is given by
u 1 = f o + f e −
u f o
u − f o
=
u f e − f 2
o − f o f e
u − f o
(2)
where f o + f e is the distance between the objective and eyepiece.
735
In BFC,
FC
BC
=
sin F ˆ
BC
sin B ˆ
FC
=
sin(π − φ)
sin 2(φ − θ)
=
sin φ
sin 2(φ − θ)
=
φ
2(φ − θ)
(3)
since the angles are small. Hence the equivalent focal length
F eq = FC =
BCφ
2(φ − θ)
=
μr
2(μ − 1)
(4)
where we have used (2), (3) and (4).
Fig. 15.23
15.14 The intensity at a distance r is given by
I =
W
4πr 2 =
100
4π × 5 2 = 0.318 W/m
2
I =
1
2
ε 0 cE
2
0
∴ E 0 =
2 I
ε 0 c
=
2 × 0.318
8.85 × 10 −12 × 3 × 10 8 = 15.48 V/m
15.15 (a) If the object distance is u then the distance v 1 at which the image is
formed by objective is given by
1
u
+
1
v 1
=
1
f o
or v 1 = −
u f o
u − f o
(1)
The distance of this image from the eyepiece, u 1 is given by
u 1 = f o + f e −
u f o
u − f o
=
u f e − f 2
o − f o f e
u − f o
(2)
where f o + f e is the distance between the objective and eyepiece.