The refractive index of a glass is 1.520 for red light and 1.525 for blue light. Let D₁ and D₂ be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then, (a) D₁ D₂

Answer: (a). For a prism, minimum deviation increases with refractive index: μ = (+D/2)/(/2). μblue = 1.525 > μred = 1.520 ⇒ D₂ > D₁ for any prism angle. So the answer is (a) D₁ < D₂.

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Prism, Dispersion & Scattering

The refractive index of a glass is $1.520$ for red light and $1.525$ for blue light. Let $D_1$ and $D_2$ be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then,

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Answer: (a)

For a prism, minimum deviation increases with refractive index: $\mu = \dfrac{\sin\frac{A+D}{2}}{\sin\frac{A}{2}}$ .
$\mu_{blue} = 1.525 > \mu_{red} = 1.520 \Rightarrow D_2 > D_1$ for any prism angle.
So the answer is (a) $D_1 < D_2$ .

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