A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle \(i\) such that it is refracted out grazing the face AD. Find the value of angle \(i\).
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For grazing refraction, the angle of refraction is \(90^\circ\), and Snell's law can be used to determine the angle of incidence in the first medium.
We use Snell's law to solve this problem, which relates the angle of incidence and refraction between two media:
\[
n_1 \sin \theta_1 = n_2 \sin \theta_2
\]
Where:
- \(n_1\) and \(n_2\) are the refractive indices of the two media,
- \(\theta_1\) is the angle of incidence in the first medium,
- \(\theta_2\) is the angle of refraction in the second medium.
Step 1: Conditions for Grazing Refraction
At the glass -liquid interface, the ray is refracted out grazing the face AD, meaning the angle of refraction at the glass -liquid interface must be \(90^\circ\).
Step 2: Using Snell’s Law at the Glass -Liquid Interface
- The refractive index of glass is \(n_{\text{glass}} = 1.5\),
- The refractive index of the liquid is \(n_{\text{liquid}} = 1.25\),
- The angle of refraction \( \theta_2 = 90^\circ \).
Using Snell's Law at the interface:
\[
n_{\text{glass}} \sin i = n_{\text{liquid}} \sin 90^\circ
\]
\[
1.5 \sin i = 1.25
\]
\[
\sin i = \frac{1.25}{1.5}
\]
\[
\sin i = \frac{5}{6}
\]
Step 3: Calculating the Angle
Now, calculate the angle of incidence \(i\):
\[
i = \sin^{ -1} \left( \frac{5}{6} \right) \approx 56.44^\circ
\]
Thus, the angle of incidence \(i \approx 56.44^\circ\).
