From the figure given below, the refractive index of medium B with respect to medium A (\(_{A}\mu_{B}\)) is:
Show 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 \(\frac{\sin(i)}{\sin(r)}\).
Step 1: Understanding the Question:
The question asks to find the refractive index of medium B relative to medium A, using the angles of incidence and refraction shown in the diagram. Step 2: Key Formula or Approach:
We will use Snell's Law of refraction, which relates the angles of incidence and refraction to the refractive indices of the two media. Snell's Law is stated as:
\[ n_1 \sin(i) = n_2 \sin(r) \]
where:
\(n_1\) = refractive index of the first medium (medium A, denoted as \(n_A\)).
\(n_2\) = refractive index of the second medium (medium B, denoted as \(n_B\)).
\(i\) = angle of incidence.
\(r\) = angle of refraction.
The refractive index of medium B with respect to medium A is given by \(_{A}\mu_{B} = \frac{n_B}{n_A}\). Step 3: Detailed Explanation:
From the given figure:
The light ray travels from Medium A to Medium B.
The angle of incidence (\(i\)) is the angle between the incident ray and the normal (N). From the diagram, \(i = 45^\circ\).
The angle of refraction (\(r\)) is the angle between the refracted ray and the normal (N'). From the diagram, \(r = 30^\circ\).
Applying Snell's Law:
\[ n_A \sin(i) = n_B \sin(r) \]
Substituting the given values:
\[ n_A \sin(45^\circ) = n_B \sin(30^\circ) \]
We need to find the refractive index of medium B with respect to medium A (\(_{A}\mu_{B}\)), which is the ratio \(\frac{n_B}{n_A}\).
Rearranging the equation to solve for \(\frac{n_B}{n_A}\):
\[ \frac{n_B}{n_A} = \frac{\sin(45^\circ)}{\sin(30^\circ)} \]
So, \(_{A}\mu_{B} = \frac{\sin(45^\circ)}{\sin(30^\circ)}\). Step 4: Final Answer:
Comparing our result with the given options, the correct choice is (A).
