Question:
A layer of oil floats on water. A ray of light strikes the oil-water interface at an angle of incidence \(30^\circ\). The refractive indices of oil and water are 1.5 and 1.3 respectively. The angle made by the ray in water is:
A layer of oil floats on water. A ray of light strikes the oil-water interface at an angle of incidence \(30^\circ\). The refractive indices of oil and water are 1.5 and 1.3 respectively. The angle made by the ray in water is:
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In multilayer refraction, always apply Snell’s law sequentially at each interface.
Updated On: Jul 18, 2026
- \(\sin^{-1}\left(\frac{1}{2}\right)\) rad
- \(\sin^{-1}\left(\frac{1.3}{1.5}\right)\) rad
- \(\sin^{-1}\left(\frac{1}{2.6}\right)\) rad
- \(\sin^{-1}\left(\frac{2}{3}\right)\) rad
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The Correct Option is C
Solution and Explanation
Step 1: Understanding refraction at layered media.
Light passes from air → oil → water, so refraction occurs at two interfaces. We apply Snell’s law step-by-step at each boundary.
Step 2: Refraction from air to oil.
Using Snell’s law: \[ 1 \cdot \sin 30^\circ = 1.5 \sin r_1 \] \[ \frac{1}{2} = 1.5 \sin r_1 \] \[ \sin r_1 = \frac{1}{3} \]
Step 3: Refraction from oil to water.
Again applying Snell’s law: \[ 1.5 \sin r_1 = 1.3 \sin r_2 \] Substitute \( \sin r_1 = \frac{1}{3} \): \[ 1.5 \cdot \frac{1}{3} = 1.3 \sin r_2 \]
Step 4: Simplification.
\[ 0.5 = 1.3 \sin r_2 \] \[ \sin r_2 = \frac{0.5}{1.3} = \frac{1}{2.6} \]
Step 5: Final angle in water.
Thus, \[ r_2 = \sin^{-1}\left(\frac{1}{2.6}\right) \]
Step 6: Final conclusion.
Hence, the angle in water is: \[ \boxed{\sin^{-1}\left(\frac{1}{2.6}\right)} \]
Light passes from air → oil → water, so refraction occurs at two interfaces. We apply Snell’s law step-by-step at each boundary.
Step 2: Refraction from air to oil.
Using Snell’s law: \[ 1 \cdot \sin 30^\circ = 1.5 \sin r_1 \] \[ \frac{1}{2} = 1.5 \sin r_1 \] \[ \sin r_1 = \frac{1}{3} \]
Step 3: Refraction from oil to water.
Again applying Snell’s law: \[ 1.5 \sin r_1 = 1.3 \sin r_2 \] Substitute \( \sin r_1 = \frac{1}{3} \): \[ 1.5 \cdot \frac{1}{3} = 1.3 \sin r_2 \]
Step 4: Simplification.
\[ 0.5 = 1.3 \sin r_2 \] \[ \sin r_2 = \frac{0.5}{1.3} = \frac{1}{2.6} \]
Step 5: Final angle in water.
Thus, \[ r_2 = \sin^{-1}\left(\frac{1}{2.6}\right) \]
Step 6: Final conclusion.
Hence, the angle in water is: \[ \boxed{\sin^{-1}\left(\frac{1}{2.6}\right)} \]
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