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JEE Main 17 March 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 28
Total: 90
The image of an object placed in air formed by a convex refracting surface is at a distance of
10
m
behind the surface. The image is real and is at
2
3
of the distance of the object from the surface .The wavelength of light inside the surface is
2
3
times the wavelength in air. The radius of the curved surface is
x
13
m
. The value of
x
is
[17 Mar 2021 Shift 2]
Your Answer:
Validate
Solution:
Given,
Image distance,
v
=
10
m
Object distance,
u
=
(
3
2
)
×
10
−
15
m
and
n
2
=
3
2
n
1
(
∵
n
∝
1
λ
)
Using the lens Maker's formula,
n
2
v
−
n
1
u
=
(
n
2
−
n
1
R
)
Substituting the values in the above equation, we get
3
2
n
1
10
−
n
1
−
15
=
(
3
2
n
1
−
n
1
R
)
⇒
R
=
30
13
m
The radius of the curved surface is
30
13
m
.
Comparing with
x
13
, we get
x
=
30
© examsnet.com
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