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# The refractive index of 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 :A. D$_{1}$< D$_{2}$ B. D$_{1}$ = D$_{2}$ C. D$_{1}$ can be less than or greater than D$_{2}$ depending upon the angle of prismD. D$_{1}$ > D$_{2}$

Last updated date: 13th Jun 2024
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Hint: D$_{1}$ and D$_{2}$ both are the angles of minimum deviation for red and blue light respectively. We need to apply this fact into the formula for the angle of minimum deviation. Divide both the angle of minimum deviation for D$_{1}$ and D$_{2}$

D$_{1}$ and D$_{2}$ is the angle of minimum deviation for both red and blue light.
RI(Refractive Index) of glass for red light is 1.502.
RI(Refractive Index) of glass for Blue light is 1.525.
The angle of minimum deviation is given by:
D=A($\mu$ -1).
Now,
$\dfrac{{{D}_{1}}}{{{D}_{2}}}=\dfrac{A({{\mu }_{R}}-1)}{A({{\mu }_{B}}-1)}$
$\dfrac{{{D}_{1}}}{{{D}_{2}}}=\dfrac{({{\mu }_{R}}-1)}{({{\mu }_{B}}-1)}$
$\dfrac{{{D}_{1}}}{{{D}_{2}}}=\dfrac{({{\mu }_{R}}-1)}{({{\mu }_{B}}-1)}$
Now since,
${{\mu }_{B}}>{{\mu }_{R}}$
$\dfrac{{{D}_{1}}}{{{D}_{2}}}<1$
${{D}_{1}}<{{D}_{2}}$ (Answer)
Hence, option A is the correct option.

If you solve the equation as a whole then the value will come in terms of D$_{1}$ or D$_{2}$.Whenever greater than less than comparison questions are given, always apply the greater value, lesser value rule. The $\mu$ in the equation of angle of minimum deviation stands for the refractive index.