Cxamples 1. If in at prism, \( \mathrm{A}=60^{\circ} \) and if \( \mathrm{m}=1.52 \), calculate the minimum deviation of a ray of light that posses through it.

1. In a prism, the minimum deviation \( \delta_{\text{min}} \) is given by the formula \[ \delta_{\text{min}} = 2\arcsin\left(n \sin\frac{A}{2}\right) - A \] where - \( A \) is the prism’s apex angle, - \( n \) is the refractive index of the prism material. 2. Given: \[ A = 60^\circ,\quad n = 1.52. \] 3. Calculate \( \sin\frac{A}{2} \): Since \( \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \) and \[ \sin 30^\circ = 0.5, \] we have \[ n \sin \frac{A}{2} = 1.52 \times 0.5 = 0.76. \] 4. Compute the value of \( \arcsin(0.76) \) (in degrees): \[ \arcsin(0.76) \approx 49.46^\circ. \] 5. Substitute into the formula for \( \delta_{\text{min}} \): \[ \delta_{\text{min}} = 2 \times 49.46^\circ - 60^\circ \approx 98.92^\circ - 60^\circ = 38.92^\circ. \] 6. Therefore, the minimum deviation of the ray is approximately \[ \delta_{\text{min}} \approx 38.92^\circ. \]