If a thin prism of glass is dipped into water then minimum deviation (with respect to air) of light produced by prism will be left $\left( \ _{a}\mu_{g} = \frac{3}{2}\ and\ _{a}\mu_{w} = \frac{4}{3} \right)$

Question

Dr. Abubacker · Accepted Answer

(b) $\frac{\delta_{a}}{\delta_{\omega}} = \frac{(_{a}\mu_{g} - 1)}{(_{\omega}\mu_{g} - 1)} = \frac{\left( \frac{3}{2} - 1 \right)}{\left( \frac{3/2}{4/3} - 1 \right)} = 4 \Rightarrow \delta_{\omega} = \frac{\delta_{a}}{4}$

Q. If a thin prism of glass is dipped into water then minimum deviation (with respect to air) of light produced by prism will be left $\left( \ _{a}\mu_{g} = \frac{3}{2}\ and\ _{a}\mu_{w} = \frac{4}{3} \right)$

Explanation

Reviewed by Dr. Abubacker

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