Question

The angle of minimum deviation for a thin prism with respect to air and when dipped in water will be: ${(}_a{\mu }_g=\frac{3}{2}$ ${}_a{\mu }_w=\frac{4}{3})$

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B $\frac{1}{4}$

We know, if $\delta$ be the angle of minimum deviation, $\mu$ be the R.I. and $A$ be the angle of prism, then,

$\delta =(\mu -1)A$

Given that, refractive index of glass, ${\mu }_g=\frac{3}{2}$, and refractive index of water, ${\mu }_w=\frac{4}{3}$.

We know, refractive index of air is ${\mu }_a=1$.

So, here, if ${\delta }_a$ and ${\delta }_w$ be the angle of minimum deviation in air and water respectively, then,

The angle of minimum deviation for a thin prism with respect to air and when dipped in water will be,

$\frac{{\delta }_w}{{\delta }_a}=\frac{\left({\mu }_g/{\mu }_w\right)-1}{\left({\mu }_g/{\mu }_a\right)-1}=\frac{\left(\frac{3}{2}/\frac{4}{3}\right)-1}{\left(\frac{3}{2}/1\right)-1}=\frac{\left(9/8\right)-1}{1/2}=\frac{1}{4}$

Therefore, correct answer is - (B) $\frac{1}{4}$