A ray of light falls on a normally on a refracting face of a prism. Find the angle of prism if the ray just fails to emerge from the prism `(mu=3//2).`

Explanation

To solve the problem of finding the angle of the prism when a ray of light falls normally on its refracting face and just fails to emerge, follow these steps:

### Step 1: Understand the Given Information
We have a prism with a refractive index (μ) of \( \frac{3}{2} \). A ray of light is incident normally on one of the refracting faces of the prism.

### Step 2: Draw a Diagram
Draw a diagram of the prism. Label the angle of the prism as \( A \). Since the ray falls normally, the angle of incidence \( i \) at the first face is \( 0^\circ \). The ray will refract into the prism.

### Step 3: Determine the Angle of Refraction
Since the ray just fails to emerge from the prism, it grazes along the second face. This means that the angle of refraction \( r \) at the second face is \( 90^\circ \).

### Step 4: Apply Snell's Law
Using Snell's Law at the second face of the prism, we have:
\[ \mu \cdot \sin(i) = n \cdot \sin(r) \]
Here, \( \mu \) is the refractive index of the prism, \( n \) is the refractive index of air (which is 1), and \( r = 90^\circ \).

### Step 5: Substitute Known Values
Substituting the known values into Snell's Law:
\[ \frac{3}{2} \cdot \sin(A) = 1 \cdot \sin(90^\circ) \]
Since \( \sin(90^\circ) = 1 \), we can simplify this to:
\[ \frac{3}{2} \cdot \sin(A) = 1 \]

### Step 6: Solve for \( \sin(A) \)
Rearranging the equation gives:
\[ \sin(A) = \frac{2}{3} \]

### Step 7: Find the Angle \( A \)
Now, we can find the angle \( A \) using the inverse sine function:
\[ A = \sin^{-1}\left(\frac{2}{3}\right) \]

### Final Answer
Thus, the angle of the prism \( A \) is:
\[ A = \sin^{-1}\left(\frac{2}{3}\right) \]

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