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.

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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  
   $$
   \delta_{\text{min}} \approx 38.92^\circ.
   $$
The minimum deviation of the ray is approximately $$ 38.92^\circ $$.

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.

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