Question

From the figure given below, the refractive index of medium B with respect to medium A (\(_{A}\mu_{B}\)) is:

• A. \(\frac{Sin45^\circ}{Sin30^\circ}\)
• B. \(\frac{Sin30^\circ}{Sin45^\circ}\)
• C. \(\frac{Sin45^\circ}{Sin60^\circ}\)
• D. \(\frac{Sin60^\circ}{Sin45^\circ}\)

Hint:

Snell's Law can be remembered as \(n_{incident} \sin(i) = n_{refracted} \sin(r)\). The relative refractive index \(_{1}\mu_{2}\) is always \(\frac{\sin(i)}{\sin(r)}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Snell's Law, the refractive index of the second medium with respect to the first medium is the ratio of the sine of the angle of incidence to the sine of the angle of refraction.
Critical Note: The angles must be measured between the ray and the normal.
Step 2: Key Formula or Approach:
\[ _A\mu_B = \frac{\sin i}{\sin r} \]
Step 3: Detailed Explanation:
From the diagram, the angle between the incident ray and the interface in Medium A is \( 45^\circ \).
The normal \( NN' \) is perpendicular to the interface.
Angle of incidence \( i = 90^\circ - 45^\circ = 45^\circ \).
In Medium B, the angle between the refracted ray and the interface is \( 30^\circ \).
Angle of refraction \( r = 90^\circ - 30^\circ = 60^\circ \).
Substituting into the formula:
\[ _A\mu_B = \frac{\sin 45^\circ}{\sin 60^\circ} \]
Step 4: Final Answer:
The correct expression is \( \frac{\sin 45^\circ}{\sin 60^\circ} \).