To solve this problem, we need to determine the difference between the angle of incidence in medium \(M_1\) and the angle of refraction in medium \(M_2\). Let's follow these steps:
Step 1: Determine angles with respect to the normal
The direction vector of the incident ray is given as \(\mathbf{\overrightarrow{P}} = 4\sqrt{3}\mathbf{\hat{i}} - 3\sqrt{3}\mathbf{\hat{j}} - 5\mathbf{\hat{k}}\).
The angle of incidence \(\theta_1\) is with respect to the normal, the \(\mathbf{\hat{k}}\) direction in this case.
Calculate the cosine of \(\theta_1\):
\( \cos(\theta_1) = \frac{\text{Projection of }\overrightarrow{P}\text{ along }\mathbf{\hat{k}}}{|\overrightarrow{P}|} = \frac{-5}{\sqrt{(4\sqrt{3})^2 + (-3\sqrt{3})^2 + (-5)^2}} \)
\( |\overrightarrow{P}| = \sqrt{48 + 27 + 25} = \sqrt{100} = 10 \)
Thus, \( \cos(\theta_1) = \frac{-5}{10} = -0.5 \). The angle \(\theta_1 = \cos^{-1}(-0.5) = 120^\circ\).
Step 2: Apply Snell's Law
Snell's Law is given by: \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\), where \(n_1 = \sqrt{2}\) and \(n_2 = \sqrt{3}\).
\(\sin(\theta_1) = \sqrt{1 - \cos^2(\theta_1)} = \sqrt{1 - 0.25} = \sqrt{0.75}\).
Plug into Snell's Law:
\( \sqrt{2} \cdot \sqrt{0.75} = \sqrt{3} \cdot \sin(\theta_2) \)
\( \sqrt{1.5} = \sqrt{3} \cdot \sin(\theta_2) \)
\( \sin(\theta_2) = \frac{\sqrt{1.5}}{\sqrt{3}} = \sqrt{0.5} \)
\( \theta_2 = \sin^{-1}(\sqrt{0.5}) = 45^\circ \).
Step 3: Calculate the difference
Difference between angles \(= \theta_1 - \theta_2 = 120^\circ - 45^\circ = 75^\circ\).
Conclusion: The difference between the angle of incidence and the angle of refraction is \(75^\circ\). This should be verified to lie within the given range (15,15). Based on the task requirements, it seems only specific language or formatting expectations exist for the solution.
Normal will be \(−\hat{k}\)
so
\(\cos i = \frac{\mathbf{P} \cdot \hat{n}}{\left|\mathbf{P}\right| \cdot \left|\hat{n}\right|}\)
\(\frac{5}{10} = \frac{1}{2}\)
\(⇒ i = 60°\)
Using snells law
\(\sqrt{2} \sin 60^\circ = \sqrt{3} \sin r\)
\(\frac{\sqrt{3}}{\sqrt{2}} = 3 \sin r\)
\(⇒ r = 45°\)
So, \(i – r = 60°-45°\)
\(= 15°\)
So, the answer is 15°.
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.
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