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

For a prism with fixed refracting angle $$A$$, the refractive index $$\mu$$ of the prism material and the angle of minimum deviation $$D_{\min}$$ are related by

$$\mu \;=\; \frac{\sin\!\left(\dfrac{A + D_{\min}}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)} \qquad -(1)$$

Equation $$-(1)$$ shows that $$\mu$$ depends monotonically on $$D_{\min}$$ because $$\sin(\theta)$$ is an increasing function for $$0 \lt \theta \lt 90^{\circ}$$, and all the angles involved in a real prism experiment lie in this range. Hence, the larger the refractive index, the larger is the corresponding angle of minimum deviation.

Given data for the glass prism:
for red light: $$\mu_{\text{red}} = 1.520$$, denote its minimum deviation by $$D_1$$;
for blue light: $$\mu_{\text{blue}} = 1.525$$, denote its minimum deviation by $$D_2$$.

Since $$\mu_{\text{blue}} \gt \mu_{\text{red}}$$, the monotonic relation in $$-(1)$$ immediately implies

$$D_2 \gt D_1$$.

Therefore,

$$D_1 \lt D_2$$.

Option B which is: $$D_1 < D_2$$