In a compound microscope the objective and eye piece have focal length 0.95 cm respectively and are kept at a distance 20 cm. The final image is formed at a distance of 25 cm from an eye piece. Calculate the position of the object and magnification
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Answer:
f
0
=0.95cm
f
e
=5cm
L=20cm
v
e
=25cm
U
0
=? M=?
L
D
=v
0
+
D+fe
Dfe
20=v
0
+
30
6
25×
5
6
95
=v
0
95
6
−
v
0
1
=
95
100
−
v
0
1
=
95
94
v
0
=−
94
95
cm
M
D
=
v
0
v
0
(1+
fe
D
)
M
D
=
94
95
6
95
(1+
5
25
)
=
6×98
95×94
×6
=94
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