the focal length of the objective and of the eyepiece of a compound microscope are f0 and fe respectively if the areas that you plan and d the least distance of distinct vision then it's angular magnification what happened when the image is formed an infinity
Answers
Explanation:
Given that,
Focal length of objective f
o
=1cm
Focal length of eyepiece f
e
=5cm
Distance of the object u
o
=1.1cm
Distinct vision = D
Now,
The magnifying power of a compound microscope is the product of the linear magnification of the objective and the magnifying power of the eyepiece.
M.P=M
o
×M
e
M.P=(
v
e
v
0
)×(
u
0
D
).....(I)
Now, for the objective lens
f
1
o
=
v
o
1
−
u
o
1
v
o
=
u
o
+f
o
u
o
f
o
u
o
v
o
=
u
o
+f
o
f
o
Now, put the value in equation (I)
M.P=(
u
o
+f
o
f
o
)×(
u
o
D
)
Now, because u
o
>f
o
So,
M.P=−(
u
o
+f
o
f
o
)×(
u
o
D
)
Now, the distance of distinct vision, D may be taken as 25 cm.
So,
M.P=−(
u
o
+f
o
f
o
)×(
u
o
D
)
M.P=−(
1.1+1
1
)×
5
25
M.P=−50
The negative sign indicates that the image is inverted
Hence, the magnifying power is 50