How does the angle of minimum deviation change of a glass prism of refractive index change if it is immersed in water?

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Solution

We know
$μ = \frac{sin (\frac{A + δ m i n}{2})}{sin (\frac{A}{2})}$ for a glam scab.
$δ m i n$ : Angle of minimum deviation
$\Rightarrow δ m i n = 2 (s i n^{-} (μ sin A / 2) - A)$
$μ$ ; Refractive index of gean - air interface
so , immersing the prism in water $μ$ decrease and so $δ$ min also decreases.

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How does the angle of minimum deviation change of a glass prism of refractive index change if it is immersed in water?

We know μ=sin(A+δmin2)sin(A2) for a glam scab.δmin: Angle of minimum deviation ⇒δmin=2(sin−(μsinA/2)−A)μ; Refractive index of gean - air interfaceso , immersing the prism in water μ decrease and so δ min also decreases.