Refractive index of a prism is `sqrt((7)/(3))` and the angle of prism is `60^@`. The limiting angle of incidence of a ray that will be transmitted through the prism is approximately

Options

A. `(pi)/(3)`
B. `(pi)/(6)`
C. `(2pi)/(3)`
D. `(pi)/(2)`

Explanation

To find the limiting angle of incidence for a ray that will be transmitted through a prism with a refractive index of \(\sqrt{\frac{7}{3}}\) and an angle of prism of \(60^\circ\), we can follow these steps:

### Step 1: Understand the Geometry of the Prism
The angle of the prism is given as \(A = 60^\circ\). When light enters the prism, it refracts at the first surface and again at the second surface. Let \(r_1\) be the angle of refraction at the first surface, and \(r_2\) be the angle of refraction at the second surface. According to the prism's geometry, we have:
\[ r_1 + r_2 = A = 60^\circ \]

### Step 2: Apply Snell's Law at the First Surface
Using Snell's Law at the first surface, we have:
\[ \sin i = \mu \sin r_1 \]
where \(i\) is the angle of incidence and \(\mu\) is the refractive index of the prism, given as \(\mu = \sqrt{\frac{7}{3}}\).

### Step 3: Express \(r_1\) in Terms of \(r_2\)
From the geometry of the prism, we can express \(r_1\) as:
\[ r_1 = 60^\circ - r_2 \]
Substituting this into Snell's Law gives us:
\[ \sin i = \mu \sin(60^\circ - r_2) \]

### Step 4: Consider the Critical Angle
For total internal reflection to occur at the second surface, \(r_2\) must be less than or equal to the critical angle \(\theta_c\). The critical angle can be calculated using:
\[ \sin \theta_c = \frac{1}{\mu} \]
Calculating \(\theta_c\):
\[ \sin \theta_c = \frac{1}{\sqrt{\frac{7}{3}}} = \sqrt{\frac{3}{7}} \]
Thus, the critical angle \(\theta_c\) can be found using:
\[ \theta_c \approx \arcsin\left(\sqrt{\frac{3}{7}}\right) \approx 40.9^\circ \]

### Step 5: Set \(r_2\) to Its Maximum Value
To find the limiting angle of incidence, we set \(r_2\) to its maximum value, which is equal to the critical angle:
\[ r_2 = \theta_c \approx 40.9^\circ \]

### Step 6: Substitute \(r_2\) Back into the Equation
Now substitute \(r_2\) back into the equation for \(r_1\):
\[ r_1 = 60^\circ - 40.9^\circ = 19.1^\circ \]
Now, substitute \(r_1\) into Snell's Law:
\[ \sin i = \mu \sin(19.1^\circ) \]

### Step 7: Calculate \(\sin i\)
Substituting the value of \(\mu\):
\[ \sin i = \sqrt{\frac{7}{3}} \sin(19.1^\circ) \]
Calculating \(\sin(19.1^\circ)\):
\[ \sin(19.1^\circ) \approx 0.327 \]
Thus,
\[ \sin i \approx \sqrt{\frac{7}{3}} \cdot 0.327 \approx 0.5 \]

### Step 8: Find the Minimum Angle of Incidence
Finally, we find the angle of incidence \(i\):
\[ i = \arcsin(0.5) = 30^\circ \]

### Final Answer
The limiting angle of incidence of a ray that will be transmitted through the prism is approximately \(30^\circ\).

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