Question

Refractive index of water w.r.t. air is 1.33. What is the refractive index of air w.r.t. water ?

• A. \(0.75\)
• B. \(0.50\)
• C. \(75.0\)
• D. \(0.25\)

Hint:

The refractive index of medium 'a' with respect to medium 'b' (\(_bn_a\)) is the reciprocal of the refractive index of medium 'b' with respect to medium 'a' (\(_an_b\)). So, \(_bn_a = 1 / _an_b\). If you know \(n_{\text{water w.r.t. air}} = 1.33 \approx 4/3\), then \(n_{\text{air w.r.t. water}}\) will be \(1 / (4/3) = 3/4 = 0.75\).

Answer 1

The Correct Option is A

Concept: The refractive index of medium 2 with respect to medium 1 (\(n_{21}\) or \(^1n_2\)) is related to the refractive index of medium 1 with respect to medium 2 (\(n_{12}\) or \(^2n_1\)) by the principle of reversibility of light. Step 1: Understanding the notation and given information "Refractive index of water w.r.t. air" can be written as \(n_{\text{water, air}}\) or \(_{\text{air}}n_{\text{water}}\). Given: \(_{\text{air}}n_{\text{water}} = 1.33\). This means when light travels from air into water, its speed changes by a factor related to 1.33, or \( \frac{\text{speed of light in air}}{\text{speed of light in water}} = 1.33 \). Step 2: What needs to be found We need to find the "refractive index of air w.r.t. water". This can be written as \(n_{\text{air, water}}\) or \(_{\text{water}}n_{\text{air}}\). This would represent \( \frac{\text{speed of light in water}}{\text{speed of light in air}} \). Step 3: Applying the principle of reversibility The principle of reversibility states that if a ray of light, after suffering any number of reflections and/or refractions, has its final path reversed, it travels back along its entire original path. This leads to the relationship between \(_{\text{air}}n_{\text{water}}\) and \(_{\text{water}}n_{\text{air}}\): \[ _{\text{water}}n_{\text{air}} = \frac{1}{_{\text{air}}n_{\text{water}}} \] In general, for any two media 1 and 2: \[ n_{12} = \frac{1}{n_{21}} \] Step 4: Substitute the given value and calculate We are given \(_{\text{air}}n_{\text{water}} = 1.33\). So, the refractive index of air with respect to water is: \[ _{\text{water}}n_{\text{air}} = \frac{1}{1.33} \] The value \(1.33\) is approximately \(4/3\). \[ _{\text{water}}n_{\text{air}} = \frac{1}{4/3} = \frac{3}{4} \] Now, convert the fraction \(3/4\) to a decimal: \[ \frac{3}{4} = 0.75 \] Step 5: Check with options The calculated value is \(0.75\), which matches option (1). Therefore, the refractive index of air w.r.t. water is \(0.75\).