Question

A glass prism whose refracting angle is 72° and refractive index 1.33. Calculate the angle of minimum deviation for the parallel beam passing through the prism.
( given: sin36°=0.5878, sin47.2°=0.7336)​

Answer 1

Answer:

$\boxed{\mathfrak{Minimum \ deviation \ (\delta_m) = 22.4 \degree}}$

Given:

Refractive angle (A) = 72°

Refractive index (μ) = 1.33

sin36° = 0.5878

sin47.2° = 0.7336

Explanation:

From Prism formula we have:

$\boxed{ \bold{ \mu = \dfrac{sin( \dfrac{A + \delta_m}{2} )}{sin \dfrac{A}{2} } }}$

$\rm \delta_m \longrightarrow$ Minimum deviation

By substituting values in the equation we get:

$\rm \implies 1.33 = \dfrac{sin( \dfrac{72 \degree + \delta_m}{2} )}{sin \dfrac{72 \degree}{2} } \\ \\ \rm \implies 1.33 = \dfrac{sin( \dfrac{72 \degree + \delta_m}{2} )}{sin36 \degree} \\ \\ \rm \implies sin( \dfrac{72 \degree + \delta_m}{2} ) = 0.5878 \times 1.33 \\ \\ \rm \implies sin( \dfrac{72 \degree + \delta_m}{2} ) = 0.781774 \\ \\ \rm \implies sin( \dfrac{72 \degree + \delta_m}{2} ) \approx sin47.2 \degree \\ \\ \rm \implies \dfrac{72 \degree + \delta_m}{2} \approx 47.2 \\ \\ \rm \implies \delta_m \approx 2 \times 47.2 \degree - 72 \degree \\ \\ \rm \implies \delta_m \approx 94.4 \degree - 72 \degree \\ \\ \rm \implies \delta_m \approx 22.4 \degree$