At the interface between two materials having refractive indices \( n_1 \) and \( n_2 \), the critical angle for reflection of an EM wave is \( \theta_c \). The \( n_1 \) material is replaced by another material having refractive index \( n_3 \), such that the critical angle at the interface between \( n_1 \) and \( n_3 \) materials is \( \theta_{c3} \). If \( n_1 > n_2 > n_3 \), \( \frac{n_2}{n_3} = \frac{2}{5} \), and \( \sin \theta_{c2} - \sin \theta_{c1} = \frac{1}{2} \), then \( \theta_{c1} \) is:
Show Hint
$\sin^{-1} \left( -\frac{5}{6} \right)$
- \( \sin^{-1} \left( \frac{2}{3n_1} \right) \)
- \( \sin^{-1} \left( \frac{1}{3n_1} \right) \)
- \( \sin^{-1} \left( \frac{1}{6n_1} \right) \)
The Correct Option is A
Approach Solution - 1
To solve this problem, we need to clearly understand the concept of the critical angle and how it relates to the refractive indices of the materials involved:
- Understanding Critical Angle:
- The critical angle \(\theta_c\) is the angle of incidence above which total internal reflection occurs when light passes from a medium with a higher refractive index to a medium with a lower refractive index.
- The relationship between the critical angle and refractive indices is given by Snell's Law at the critical angle: \[ \sin \theta_c = \frac{n_2}{n_1} \] where \(n_1\) is the refractive index of the first medium and \(n_2\) is that of the second medium.
- Given Problem Information:
- Refractive indices: \( n_1 > n_2 > n_3 \)
- \(\frac{n_2}{n_3} = \frac{2}{5}\)
- \(\sin \theta_{c2} - \sin \theta_{c1} = \frac{1}{2}\)
- Calculating Critical Angles:
- First, note that \(\sin \theta_{c2} = \frac{n_3}{n_1}\) and \(\sin \theta_{c1} = \frac{n_2}{n_1}\).
- Using the given condition \( \sin \theta_{c2} - \sin \theta_{c1} = \frac{1}{2} \): \[ \frac{n_3}{n_1} - \frac{n_2}{n_1} = \frac{1}{2} \] \[ \frac{n_3 - n_2}{n_1} = \frac{1}{2} \] \[ n_3 - n_2 = \frac{n_1}{2} \]
- Since \(\frac{n_2}{n_3} = \frac{2}{5}\), it implies \(n_2 = \frac{2}{5}n_3\).
- Solving for \(\theta_{c1}\):
- Substitute \(n_3 = \frac{5}{2}n_2\) in the equation \(n_3 - n_2 = \frac{n_1}{2}\): \[ \frac{5}{2}n_2 - n_2 = \frac{n_1}{2} \] \[ \frac{3}{2}n_2 = \frac{n_1}{2} \] \[ n_1 = 3n_2 \]
- Now, using \(\sin \theta_{c1} = \frac{n_2}{n_1}\): \[ \sin \theta_{c1} = \frac{n_2}{3n_2} = \frac{1}{3} \]
- Conclusion:
- The value of \(\sin \theta_{c1}\) calculated does not match any provided option for \(\theta_{c1}\). Hence, an error may have occurred in the given options or conditions.
- Revisiting the particular options, it appears that there has been a mismatch, but the calculation of solving general condition shows that provided result \(\sin^{-1} \left( -\frac{5}{6} \right)\) cannot be correct since sine function cannot have negative values greater than -1. Correctly calculated discrete values like \(\frac{1}{6}\) etc., imply an error in the presumption provided.
- Thus, logical reasoning indicates to revise calculation or retry implementation based on textbook standards over mis-choice inconsistencies.
Approach Solution -2
We are tasked with solving for the critical angles $ \theta_{1C} $ and $ \theta_{2C} $, and determining the relationship between the refractive indices $ n_1 $, $ n_2 $, and $ n_3 $. The solution proceeds as follows:
1. Critical Angle Relations:
The sine of the critical angle is given by:
$ \sin \theta_{1C} = \frac{n_1}{n_2} $
$ \sin \theta_{2C} = \frac{n_1}{n_3} $
2. Difference Between Sines:
We are given that:
$ \sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2} $
Substituting the expressions for $ \sin \theta_{1C} $ and $ \sin \theta_{2C} $:
$ \frac{n_1}{n_3} - \frac{n_1}{n_2} = \frac{1}{2} $
3. Simplifying the Equation:
Factor out $ n_1 $:
$ n_1 \left( \frac{1}{n_3} - \frac{1}{n_2} \right) = \frac{1}{2} $
Simplify the terms inside the parentheses:
$ \frac{1}{n_3} - \frac{1}{n_2} = \frac{n_2 - n_3}{n_2 n_3} $
Thus, the equation becomes:
$ n_1 \cdot \frac{n_2 - n_3}{n_2 n_3} = \frac{1}{2} $
4. Substituting Given Values:
From the problem, we know:
$ n_1 \left( \frac{2}{5} - 1 \right) = \frac{n_3}{2} $
Simplify $ \frac{2}{5} - 1 $:
$ \frac{2}{5} - 1 = \frac{2}{5} - \frac{5}{5} = -\frac{3}{5} $
Substitute this back into the equation:
$ n_1 \cdot \left( -\frac{3}{5} \right) = \frac{n_3}{2} $
Solve for $ n_1 $:
$ n_1 = \frac{n_3}{2} \cdot \left( -\frac{5}{3} \right) = -\frac{5}{6} n_3 $
5. Ratio of Refractive Indices:
We now determine the ratio $ \frac{n_1}{n_2} $. From earlier, we know:
$ \frac{n_1}{n_2} = \frac{-5}{6} $
6. Inverse Sine Calculation:
Finally, the critical angle $ \theta_{1C} $ is given by:
$ \theta_{1C} = \sin^{-1} \left( \frac{n_1}{n_2} \right) $
Substitute $ \frac{n_1}{n_2} = \frac{-5}{6} $:
$ \theta_{1C} = \sin^{-1} \left( -\frac{5}{6} \right) $
Final Answer:
The critical angle $ \theta_{1C} $ is: $ \sin^{-1} \left( -\frac{5}{6} \right) $
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