A ray of light is incident on an air-glass boundary at an angle \(\theta\). If the angle between the partially reflected ray and the refracted ray is 90°, c...
Question 1 Report
A ray of light is incident on an air-glass boundary at an angle \(\theta\). If the angle between the partially reflected ray and the refracted ray is 90°, calculate \(\theta\), given that the refractive index of glass is 1.50.
Answer Details
Finding the angle of incidence \( \theta \)
The reflected ray makes an angle \( \theta \) with the normal (equal to the angle of incidence), and the refracted ray makes an angle \( r \) with the normal on the other side. If the angle between the reflected ray and the refracted ray is \( 90^\circ \), then, measuring from the normal on each side,
\[ \theta + 90^\circ + r = 180^\circ \quad\Rightarrow\quad \theta + r = 90^\circ \quad\Rightarrow\quad r = 90^\circ - \theta \]
Applying Snell's law at the air-glass boundary (\( n = 1.50 \)):
\[ \sin\theta = n\sin r = n\sin(90^\circ - \theta) = n\cos\theta \]
\[ \frac{\sin\theta}{\cos\theta} = n \quad\Rightarrow\quad \tan\theta = 1.50 \]
\[ \theta = \tan^{-1}(1.50) = 56.3^\circ \]
The angle of incidence is \( \theta \approx 56.3^\circ \). (This is the polarizing, or Brewster, angle, for which \( \tan\theta = n \).)
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