The magnitude produced by a convex lens 2 write the value in both si and cgs unit .
Answers
12th
Physics
Ray Optics and Optical Instruments
Lens Formula and Magnification
The magnifications produced...
PHYSICS
The magnifications produced by a convex lens for two different positions of an object are m
1
and m
2
respectively (m
1
> m
2
). If d is the distance of separation between the two positions of the object then the focal length of the lens is
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ANSWER
Given: The magnifications produced by a convex lens for two different positions of an object are m
1
and m
2
respectively (m
1
>m
2
). If d is the distance of separation between the two positions of the object
To find the focal length of the lens
Solution:
Separation between object and image for first position,
D=v+u
where u,v are the object distance and image distance, respectively.
So, magnification
m
1
=
u
v
⟹v=um
1
....(i)
When the lens is at second position,
d=v-u
So, m
2
=
v
u
⟹u=vm
2
......(ii)
So, m
1
m
2
=1
From lens equation and using eqn (i), we get
f
1
=
v
1
+
u
1
⟹
f
1
=
um
1
1
+
u
1
⟹
f
1
=
um
1
1+m
1
⟹u=
m
1
f(m
1
+1)
.........(iii)
From lens equation and using eqn (ii), we get
f
1
=
v
1
+
u
1
⟹
f
1
=
vm
2
1
+
v
1
⟹
f
1
=
vm
2
1+m
2
⟹v=
m
2
f(m
2
+1)
.........(iv)
substituting the values of eqn(iii) and(iv) in the following equation, we get
d=v−u
⟹d=
m
2
f(m
2
+1)
−
m
1
f(m
1
+1)
⟹d=f(
m
1
m
2
m
1
(m
2
+1)−m
2
(m
1
+1)
)
⟹f=
m
1
m
2
+m
1
−m
1
m
2
−m
2
dm
1
m
2
⟹f=
m
1
−m
2
dm
1
m
2
but m
1
m
2
=1
Therefore the focal length becomes,
f=
m
1
−m
2
d
solution