Concept: The refractive index of medium 2 with respect to medium 1 (\(n_{21}\) or \(^1n_2\)) is given by the ratio of the absolute refractive index of medium 2 (\(n_2\)) to the absolute refractive index of medium 1 (\(n_1\)).
Absolute refractive index is usually measured with respect to air (or vacuum).
\[ n_{21} = \frac{n_2}{n_1} = \frac{\text{Refractive index of medium 2 w.r.t. air}}{\text{Refractive index of medium 1 w.r.t. air}} \]
This can also be expressed in terms of speeds of light: \(n_{21} = \frac{v_1}{v_2}\), where \(v_1\) is the speed of light in medium 1 and \(v_2\) is the speed of light in medium 2.
Step 1: Identify the given refractive indices (with respect to air)
Let \(n_g\) be the refractive index of flint glass with respect to air.
Let \(n_a\) be the refractive index of alcohol with respect to air.
Given:
\(n_g = 1.65\)
\(n_a = 1.36\)
Step 2: Determine what needs to be found
We need to find the refractive index of flint glass with respect to alcohol. Let this be \(n_{ga}\) (glass w.r.t. alcohol).
Here, flint glass is medium 2 and alcohol is medium 1.
So, \(n_{ga} = \frac{n_g}{n_a}\).
Step 3: Substitute the given values and calculate
\[ n_{ga} = \frac{1.65}{1.36} \]
Now, perform the division:
\[ 1.65 \div 1.36 \approx 1.2132... \]
Rounding to two decimal places, we get \(1.21\).
The refractive index of flint glass with respect to alcohol is approximately 1.21.
This matches option (3).