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Standard X
Physics
Question
For prism of refractive index
1.732
, the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is:
80
0
60
0
70
0
50
0
A
80
0
B
70
0
C
60
0
D
50
0
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Solution
Verified by Toppr
μ
=
1.732
D
m
=
A
μ
=
s
i
n
(
A
+
D
m
2
)
s
i
n
(
A
2
)
1.732
=
s
i
n
(
2
A
2
)
s
i
n
(
A
2
)
1.732
=
2
s
i
n
A
2
⋅
c
o
s
A
2
s
i
n
(
A
2
)
c
o
s
A
2
=
1.732
2
A
2
=
c
o
s
−
1
(
0.866
)
A
2
=
30
0
(or)
A
=
60
0
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