The focal length of the object and the eyepiece of a compound microscope are 1 cm
and 2.5 cm respectively and the length of the tube is 16 cm. Find the magnifying
power of the microscope, if the final image is formed at infinity
Answers
Answered by
0
Answer:
Given,
Separation between lense,L=21.7cm,
Focal length of eye piece, L=21.7cm,f
e
=2.5cm
Focal length of objective lens,f
0
=1.6cm,v
0
=19.2cm,u
0
=?
L=v
0
+f
e
Where L is the length of the microscope and f
e
e the focal length of the objective
Substituting we get:
21.7=v
0
+2.5
⇒v
0
=21.7−2.5=19.2cm
According to lens formula we know that:
f
0
1
=
v
0
1
−
u
0
1
v
0
=19.2cm,u
0
=?
1.6
1
=
19.2
1
−
u
0
1
−
u
0
1
=
1.6
1
−
19.2
1
−
u
0
1
=
16
10
−
192
10
−
u
0
1
=
192
120−10
=
192
110
u
0
=−
110
192
cm=−1.75cm
Magnification, m=
u
−v
=
1.75
−19.2
=10.97=11
Hence, linear magnification is 11
Answered by
0
Answer:
10.97 = 11
Explanation:
your answer ok ok ok ok ok ok ok
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