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JEE Mains 24-Jan-2023 Shift 1 Solved Paper
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© examsnet.com
Question : 26
Total: 90
As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is
30
cm
and refraction index of the material for both the lenses is
1.75
. Both the lenses are placed at distance of 40
cm
from each other. Due to the combination, the image of the object is formed at distance
x
=
___
cm
, from concave lens.
[24-Jan-2023 Shift 1]
Your Answer:
Validate
Solution:
1
f
1
=
(
1.75
−
1
)
(
−
1
30
)
⇒
f
1
=
−
40
cm
1
f
2
=
(
1.75
−
1
)
(
1
30
)
⇒
f
2
=
40
cm
Image from
L
1
will be virtual and on the left of
L
1
at focal length
40
cm
. So the object for
L
2
will be
80
cm
from
L
2
which is
2
f
. Final image is formed at
80
cm
from
L
2
on the right.
So
x
=
120
© examsnet.com
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