736
15 Optics
If the final image is formed at a distance v from the eyepiece then
1
v
+
1
u 1
=
1
f e
or
1
v
=
1
f e
−
u − f o
u f e − f 2
e − f o f e
or v = −
f e (u f e − f 2
o − f o f e )
f 2
o
(3)
Now u >> f e , f o
∴
v = −u
f e
f o
2
= 10
4
1
8
2
= 156.25 m
(b) M =
Height of the final image
Height of the object
=
v 1
u
×
v
u 1
−
f e
f o
= −
1
8
= −0.125
where we have used (1), (2) and (3).
∴
Height of final image = (height of the object).M
= −100 × 0.125 = −12.5 m
The negative sign indicates that the final image is virtual.
Note that the height of the image is only one-eighth of that of the
object, but the image is closer than the object by a factor of 64, so it
subtends an angle eight times large, that is, the image appears eight times
larger.
15.16 (a) The Intensity is given by
I =
power
cross-section
=
W
A
=
1000
10 −5 × 10 −4 = 10
12 W/m
2
(b) I =
1
2
ε 0 cE
2
0
E 0 =
2I
ε 0 c
=
2 × 10 12
8.85 × 10 −12 × 3 × 10 8
= 2.74 × 10
7 V/m
15 Optics
If the final image is formed at a distance v from the eyepiece then
1
v
+
1
u 1
=
1
f e
or
1
v
=
1
f e
−
u − f o
u f e − f 2
e − f o f e
or v = −
f e (u f e − f 2
o − f o f e )
f 2
o
(3)
Now u >> f e , f o
∴
v = −u
f e
f o
2
= 10
4
1
8
2
= 156.25 m
(b) M =
Height of the final image
Height of the object
=
v 1
u
×
v
u 1
−
f e
f o
= −
1
8
= −0.125
where we have used (1), (2) and (3).
∴
Height of final image = (height of the object).M
= −100 × 0.125 = −12.5 m
The negative sign indicates that the final image is virtual.
Note that the height of the image is only one-eighth of that of the
object, but the image is closer than the object by a factor of 64, so it
subtends an angle eight times large, that is, the image appears eight times
larger.
15.16 (a) The Intensity is given by
I =
power
cross-section
=
W
A
=
1000
10 −5 × 10 −4 = 10
12 W/m
2
(b) I =
1
2
ε 0 cE
2
0
E 0 =
2I
ε 0 c
=
2 × 10 12
8.85 × 10 −12 × 3 × 10 8
= 2.74 × 10
7 V/m