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IIT JEE Advanced 2010 Paper 1
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© examsnet.com
Question : 22
Total: 84
The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from
m
25
to
m
20
. The ratio
m
25
m
50
is _____.
Your Answer:
Validate
Solution:
The magnification is expressed as m =
f
f
+
u
Therefore,
m
25
m
50
=
f
+
u
50
f
+
u
25
=
20
−
50
20
−
25
=
−
30
−
5
= 6
Alternative Solution:
Using the lens formula,
1
v
+
1
u
=
1
f
, for the two positions of the object gives
1
v
1
−
1
−
25
=
1
20
;
1
v
2
−
1
−
50
=
1
20
.
From the first equation, we get
v
1
= 100 cm and the second equation gives
v
2
= 100/3 cm. Hence, the magnification is thefirst case is
m
25
=
v
1
u
1
= - 4 and that in the second case is
m
50
=
v
2
u
2
=
−
2
3
. Hence ,
m
25
m
50
= 6.
© examsnet.com
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