Question:
A ray of light is incident at \(30^\circ\) from a medium of refractive index \(2\) into a medium of refractive index \(1\). Then the angle of refraction is:
A ray of light is incident at \(30^\circ\) from a medium of refractive index \(2\) into a medium of refractive index \(1\). Then the angle of refraction is:
Show Hint
If the refracted angle becomes
\[
90^\circ,
\]
the angle of incidence is equal to the critical angle.
Updated On: Jun 24, 2026
- \(30^\circ\)
- \(60^\circ\)
- \(45^\circ\)
- \(90^\circ\)
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The Correct Option is D
Solution and Explanation
Step 1: Use Snell’s law.
According to Snell’s law, \[ n_1\sin i=n_2\sin r \] where \[ n_1=2, \quad n_2=1, \quad i=30^\circ \]
Step 2: Substitute the values.
\[ 2\sin30^\circ=1\cdot \sin r \] Since \[ \sin30^\circ=\frac{1}{2}, \] we get \[ 2\times \frac{1}{2}=\sin r \] \[ \sin r=1 \]
Step 3: Find the angle of refraction.
\[ r=90^\circ \]
Step 4: Final conclusion.
Hence, the angle of refraction is \[ \boxed{90^\circ} \]
According to Snell’s law, \[ n_1\sin i=n_2\sin r \] where \[ n_1=2, \quad n_2=1, \quad i=30^\circ \]
Step 2: Substitute the values.
\[ 2\sin30^\circ=1\cdot \sin r \] Since \[ \sin30^\circ=\frac{1}{2}, \] we get \[ 2\times \frac{1}{2}=\sin r \] \[ \sin r=1 \]
Step 3: Find the angle of refraction.
\[ r=90^\circ \]
Step 4: Final conclusion.
Hence, the angle of refraction is \[ \boxed{90^\circ} \]
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