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JEE Advanced 2019 Paper 1
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© examsnet.com
Question : 5
Total: 54
A thin convex lens is made of two materials with refractive indices
n
1
and
n
2
, as shown in figure. The radius of curvature of the left and right spherical surface are equal. f is the focal length of the lens when
n
1
=
n
2
=
n
.The focal length is
f
+
Δ
f
when
n
1
=
n
and
n
2
=
n
+
Δ
n
.Assuming
Δ
n
<
<
(
n
−
1
)
and
1
<
<
2
,the correct statement(s) is/are,
If
Δ
n
n
<
0
then
Δ
f
f
>
0
For
n
=
1.5
,
Δ
n
=
10
−
3
and
f
=
20
c
m
,the value of
|
Δ
f
|
will be 0.02 cm (rounded off to 2nd decimal place).
The relation between
∆
f
f
and
∆
n
n
remains unchanged if both the convex surfaces arereplaced by concave surface of the same radius of curvature.
|
∆
f
f
|
<
|
∆
n
n
|
Validate
Solution:
f
=
R
2
(
n
−
1
)
...........(1)
In 2nd case
1
f
+
d
f
=
n
−
1
R
+
(
n
+
d
n
)
−
1
R
1
f
+
d
f
=
1
f
+
d
n
R
Clearly,if
d
n
<
0
,R.H.S value is less than
1
f
∴
d
f
>
0
i.e (1) is correct
Option(2)
f
=
R
2
(
1.5
−
1
)
⇒
R
=
20
c
m
1
f
+
d
f
=
1
f
+
d
n
R
1
20
+
d
f
=
1
20
+
d
n
20
1
20
(
1
+
d
f
20
)
=
1
20
(
1
+
d
n
)
1
−
d
f
20
=
1
+
d
n
⇒
d
f
=
−
20
d
n
=
−
20
×
10
−
3
cm
∴
|
d
f
|
=
0.02
cm
(2) is correct
Option(3): If convex surfaces are replaced with concave surfaces, sign of R will change
1
f
(
1
+
d
f
f
)
=
1
f
+
d
n
R
=
1
f
+
d
n
2
f
(
n
−
1
)
(
f
=
R
2
(
n
−
1
)
1
R
=
1
2
f
(
n
−
1
)
)
1
−
d
f
f
f
=
1
f
[
1
+
d
n
2
n
(
1
−
1
n
)
]
−
d
f
f
=
(
d
n
n
)
1
2
(
1
−
1
n
)
As this is independent of R, (3) is correct
Option (4):
|
d
f
f
|
=
|
d
n
n
|
1
2
(
1
−
1
n
)
For the range
1
<
n
<
2
,
2
(
1
−
1
n
)
is always greater than 1 i.e,
|
d
f
f
|
>
|
d
n
n
|
∴ (4) is wrong
© examsnet.com
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