To solve this problem, we need to determine what happens to a ray of light as it passes from a medium with a given refractive index into air, given a specific angle of incidence.
The refractive index of the medium is given as \(\frac{1}{\sqrt{2}}\). When light travels from this medium to air, we are interested in the refraction or reflection behavior at the interface.
The angle of incidence is given to be \(45^\circ\) with the surface. This means the angle with the normal is also \(45^\circ\) (since angle with surface + angle with normal = 90°).
Using Snell's Law:
\(n_1 \cdot \sin(\theta_1) = n_2 \cdot \sin(\theta_2)\)
Where:
Applying the values to Snell's Law:
\(\frac{1}{\sqrt{2}} \cdot \sin(45^\circ) = 1 \cdot \sin(\theta_2)\)
Since \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\), we can write:
\(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{2} = \sin(\theta_2)\)
\(\frac{1}{2} = \sin(\theta_2)\)
The maximum value of \(\sin(\theta_2)\) is 1, which corresponds to an angle of \(90^\circ\). Therefore, for \(\sin(\theta_2) = \frac{1}{2}\), the situation requires internal reflection because the condition violates the law of refraction for the given incident angle and the calculated sine value exceeds the physical limit of total internal reflection.
The critical angle for total internal reflection can be checked by setting:
\(n_1 \cdot \sin(\theta_c) = n_2 \cdot \sin(90^\circ)\)
Solving for the critical angle \(\theta_c\):
\(\sin(\theta_c) = \frac{n_2}{n_1}\)
Since \(n_2 = 1\) and \(n_1 = \frac{1}{\sqrt{2}}\), this yields:
\(\sin(\theta_c) = \sqrt{2}\)
Clearly, this confirms that the angle requirement cannot be met physically here, indicating total internal reflection occurs.
Thus, The ray will be internally reflected.