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Section:
Mathematics
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Question : 3 of 59
Marks:
+1
,
-0
Consider two isosceles prisms 1 and 2 with prism angles
A
1
A_1
A
1
and
A
2
A_2
A
2
and refractive indices
n
1
n_1
n
1
and
n
2
n_2
n
2
, respectively, as shown in the figure. The faces
a
1
b
1
a_1 b_1
a
1
b
1
and
a
2
b
2
a_2 b_2
a
2
b
2
are parallel to each other and perpendicular to the mirror
M
M
M
. If a ray of light is incident on the face
a
1
c
1
a_1 c_1
a
1
c
1
and emerges from the face
a
2
c
2
a_2 c_2
a
2
c
2
, then the correct statement(s) is/are :
[JEE Adv 2026 P2]
If both the prisms are at minimum deviation condition, then
n
2
n
1
=
sin
(
A
1
2
)
sin
(
A
2
2
)
\dfrac{n_2}{n_1} = \dfrac{\sin\left(\dfrac{A_1}{2}\right)}{\sin\left(\dfrac{A_2}{2}\right)}
n
1
n
2
=
sin
(
2
A
2
)
sin
(
2
A
1
)
.
If prism 2 is at minimum deviation condition, then
sin
i
1
=
n
2
sin
(
A
2
2
)
\sin i_1 = n_2 \sin \left(\dfrac{A_2}{2}\right)
sin
i
1
=
n
2
sin
(
2
A
2
)
is always true.
If both the prisms 1 and 2 are thin and are at minimum deviation condition with angles of deviation
δ
m
1
\delta_{m1}
δ
m
1
and
δ
m
2
\delta_{m2}
δ
m
2
, respectively, then
θ
=
δ
m
1
2
(
n
1
−
1
)
+
δ
m
2
2
(
n
2
−
1
)
\theta = \dfrac{\delta_{m1}}{2(n_1 - 1)} + \dfrac{\delta_{m2}}{2(n_2 - 1)}
θ
=
2
(
n
1
−
1
)
δ
m
1
+
2
(
n
2
−
1
)
δ
m
2
.
If prism 1 is at minimum deviation condition, then
sin
i
2
=
n
1
sin
(
A
1
2
)
\sin i_2 = n_1 \sin \left(\dfrac{A_1}{2}\right)
sin
i
2
=
n
1
sin
(
2
A
1
)
is always true.
Validate
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