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JEE Advanced 2017 Paper 1
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© examsnet.com
Question : 6
Total: 54
For an isosceles prism of angle A and refractive index
µ
, it is found that the angle of minimum deviation
δ
m
=
A
which of the following is/are correct
At minimum deviation, the incident angle
i
1
and the refracting angle
r
1
at the first refracting surface are realted by
r
1
=
(
i
1
2
)
For this prism, the refractive index μ and the angle of prism A are related as
A
=
1
2
c
o
s
−
1
(
µ
2
)
For the angle of incidence
i
1
=
A
,the ray inside the prism is parallel to the base of the prism
For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is
i
1
=
s
i
n
−
1
A
√
4
c
o
s
2
A
2
−
1
−
c
o
s
A
Validate
Solution:
Angular deviation :
δ
=
i
1
+
i
2
−
A
For min deviation
:
i
1
=
1
2
so,
i
1
=
A
...(i)
Also
r
1
+
r
2
=
A
i.e.,
r
1
=
A
2
.....(ii)
∴
r
1
=
i
1
2
(A is correct)
By Snell's law,
s
i
n
i
1
=
µ
r
1
i.e
s
i
n
A
=
µ
s
i
n
(
A
2
)
⇒
2
c
o
s
(
A
2
)
=
µ
A
=
2
c
o
s
−
1
(
µ
2
)
Hence (B) is correct
(C) is obviously correct
For tangential emergence,
i
2
→
90
°
So,
µ
s
i
n
r
2
=
1
c
o
s
r
22
=
√
1
−
1
µ
2
Also
r
1
=
A
−
r
2
s
i
n
r
1
=
s
i
n
A
c
o
s
r
2
−
c
o
s
A
s
i
n
r
2
=
(
s
i
n
A
)
√
µ
2
−
1
µ
−
(
c
o
s
A
)
1
µ
By Snell's law on 1st surface,
s
i
n
i
2
=
µ
s
i
r
1
⇒
s
i
n
i
1
=
s
i
n
A
(
µ
2
−
1
)
1
2
−
c
o
s
A
⇒
i
1
s
i
n
−
1
(
s
i
n
A
√
4
c
o
s
2
A
2
−
1
−
c
o
s
A
)
i.e. (D) is correct
© examsnet.com
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