The refractive index of air with respect to glass is \( \frac{2}{3} \). The refractive index of diamond with respect to air is 2.4. What will be the refractive index of glass with respect to diamond?
0.625
Understanding Refractive Index Calculations
This problem requires us to calculate the refractive index of one medium with respect to another, using given relative refractive indices. The refractive index is a fundamental property of a medium that describes how light propagates through it.
Key Concepts: Refractive Index
- Absolute Refractive Index: The refractive index of a medium with respect to vacuum (or air, approximately). It is denoted by \(n\) and is given by the ratio of the speed of light in vacuum (\(c\)) to the speed of light in the medium (\(v\)), i.e., \(n = \frac{c}{v}\).
- Relative Refractive Index: The refractive index of medium 2 with respect to medium 1 is denoted by \(n_{21}\) or \(_1 n_2\). It is the ratio of the speed of light in medium 1 (\(v_1\)) to the speed of light in medium 2 (\(v_2\)), i.e., \(n_{21} = \frac{v_1}{v_2}\). It can also be expressed in terms of absolute refractive indices as \(n_{21} = \frac{n_2}{n_1}\).
Analyzing the Given Information
We are given the following:
- The refractive index of air with respect to glass is \( \frac{2}{3} \). We can write this as \( n_{ag} = \frac{2}{3} \). Using the definition \( n_{21} = \frac{n_2}{n_1} \), this means \( \frac{n_a}{n_g} = \frac{2}{3} \), where \(n_a\) is the absolute refractive index of air and \(n_g\) is the absolute refractive index of glass.
- The refractive index of diamond with respect to air is 2.4. We can write this as \( n_{da} = 2.4 \). This means \( \frac{n_d}{n_a} = 2.4 \), where \(n_d\) is the absolute refractive index of diamond. Note that \(n_{da}\) is essentially the absolute refractive index of diamond since \(n_a \approx 1\).
Goal: Find Refractive Index of Glass with Respect to Diamond
We need to find the refractive index of glass with respect to diamond, which is \( n_{gd} \). According to the definition, \( n_{gd} = \frac{n_g}{n_d} \).
Step-by-Step Calculation
We have the ratios \( \frac{n_a}{n_g} \) and \( \frac{n_d}{n_a} \). We need to find \( \frac{n_g}{n_d} \).
From the first given piece of information:
\( n_{ag} = \frac{n_a}{n_g} = \frac{2}{3} \)
We can find the ratio of the absolute refractive index of glass to that of air by taking the reciprocal:
\( \frac{n_g}{n_a} = \frac{1}{n_{ag}} = \frac{1}{2/3} = \frac{3}{2} = 1.5 \)
From the second given piece of information:
\( n_{da} = \frac{n_d}{n_a} = 2.4 \)
Now we want to find \( n_{gd} = \frac{n_g}{n_d} \). We can express this ratio by dividing the absolute refractive index of glass (relative to air) by the absolute refractive index of diamond (relative to air):
\( n_{gd} = \frac{n_g}{n_d} = \frac{n_g/n_a}{n_d/n_a} \)
Substitute the values we found:
\( n_{gd} = \frac{1.5}{2.4} \)
To simplify this fraction, we can write it as:
\( n_{gd} = \frac{15}{24} \)
Both 15 and 24 are divisible by 3:
\( n_{gd} = \frac{15 \div 3}{24 \div 3} = \frac{5}{8} \)
Now, convert the fraction to a decimal:
\( \frac{5}{8} = 5 \div 8 = 0.625 \)
So, the refractive index of glass with respect to diamond is 0.625.
Summary of the Refractive Index Calculation
We used the definition of relative refractive index in terms of absolute refractive indices to relate the given values to the required value. By finding the ratio of the absolute refractive index of glass to that of air (\(n_g/n_a\)) and the ratio of the absolute refractive index of diamond to that of air (\(n_d/n_a\)), we could calculate the ratio \(n_g/n_d\).
| Given Information | In Terms of Absolute Indices | Value |
|---|---|---|
| Refractive index of air w.r.t. glass (\(n_{ag}\)) | \( \frac{n_a}{n_g} \) | \( \frac{2}{3} \) |
| Refractive index of diamond w.r.t. air (\(n_{da}\)) | \( \frac{n_d}{n_a} \) | 2.4 |
| Calculated Values | Derivation | Value |
|---|---|---|
| Ratio of absolute index of glass to air (\(n_g/n_a\)) | \( \frac{1}{n_{ag}} = \frac{1}{2/3} \) | 1.5 |
| Refractive index of glass w.r.t. diamond (\(n_{gd}\)) | \( \frac{n_g/n_a}{n_d/n_a} = \frac{1.5}{2.4} \) | 0.625 |
Revision Table: Refractive Index Formulas
| Concept | Formula | Description |
|---|---|---|
| Absolute Refractive Index | \( n = \frac{c}{v} \) | Ratio of speed of light in vacuum to speed of light in medium. |
| Relative Refractive Index (Med 2 w.r.t Med 1) | \( n_{21} = \frac{v_1}{v_2} = \frac{n_2}{n_1} \) | Ratio of speed of light in Med 1 to speed of light in Med 2, or ratio of absolute indices. |
| Reciprocal Rule | \( n_{21} = \frac{1}{n_{12}} \) | Refractive index of Med 2 w.r.t Med 1 is reciprocal of Med 1 w.r.t Med 2. |
| Chaining Rule (using absolute indices) | \( n_{31} = \frac{n_3}{n_1} = \frac{n_3/n_a}{n_1/n_a} = \frac{n_{3a}}{n_{1a}} \) | Calculating relative index using indices relative to a common medium (like air). |
Additional Information: Refractive Index and Speed of Light
The refractive index is directly related to how much the speed of light is reduced when it enters a medium from vacuum. A higher refractive index means light travels slower in that medium. For example, the refractive index of diamond (2.4) is much higher than that of glass (around 1.5), meaning light travels significantly slower in diamond than in glass. This difference in speed is what causes light to bend when it passes from one medium to another, a phenomenon called refraction.
When light goes from a medium with a lower refractive index (like air) to a medium with a higher refractive index (like glass or diamond), it bends towards the normal (the line perpendicular to the surface). When it goes from a higher refractive index medium to a lower one, it bends away from the normal.
Understanding refractive indices is crucial in optics, for designing lenses, prisms, and other optical instruments. The dispersion of light into its constituent colors by a prism is also explained by the fact that the refractive index of the prism material is slightly different for different wavelengths (colors) of light.
Important Questions from Ray Optics and Optical Instruments
A Convex mirror produces the magnification 1/3 and 1/4 when the object is placed at the points P and Q in front of the mirror.
Which of the following statements are correct?
- A. The saturation current is constant with collector plate potential for different frequencies of incident radiation.
- B. The saturation current is different with collector plate potential for different frequencies of incident radiation.
- C. The saturation current is different with collector plate potential for different intensity of incident radiation.
- D. The saturation current is constant with collector plate potential for different intensity of incident radiation.
- E. Below threshold frequency, no photoelectrons are emitted.
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