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Problem
Problem
Assertion : Minimum deviation by an equilateral prism of refractive index
2
\sqrt{2}
2
is
30
∘
30^\circ
3
0
∘
. Reason : It is from the relation,
μ
=
sin
(
A
+
δ
m
2
)
sin
(
A
2
)
\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}
μ
=
s
i
n
(
2
A
)
s
i
n
(
2
A
+
δ
m
)
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