The refracting angle of a glass prism is `30^@.` A ray is incident onto one of the faces perpendicular to it. Find the angle `delta` between the incident ray and the ray that leaves the prism. The refractive index of glass is `mu=1.5.`

Explanation

To solve the problem step by step, we will follow the principles of refraction and the properties of a prism.

### Step 1: Identify the Given Values
- Refracting angle of the prism (A) = 30°
- Refractive index of glass (μ) = 1.5
- The ray is incident perpendicularly on one face of the prism.

### Step 2: Understand the Incident Ray
Since the ray is incident perpendicularly to one of the faces of the prism, the angle of incidence (I1) is 0°.

### Step 3: Determine the Angle of Refraction at the First Face
According to the law of refraction (Snell's Law):
\[ \mu = \frac{\sin I_2}{\sin R_1} \]
Where:
- I2 is the angle of refraction at the first face.
- R1 is the angle of refraction at the first face.

Since the angle of incidence (I1) is 0°, the angle of refraction (R1) will also be 0° (as they are coaxial).

### Step 4: Relate the Angles of Refraction
The angle of refraction at the second face (R2) can be related to the refracting angle of the prism (A):
\[ R_2 + R_1 = A \]
Since R1 = 0°:
\[ R_2 = A = 30° \]

### Step 5: Apply Snell's Law at the Second Face
Now we apply Snell's Law at the second face:
\[ \mu = \frac{\sin I_2}{\sin R_2} \]
Where R2 = 30°.
Thus:
\[ 1.5 = \frac{\sin I_2}{\sin 30°} \]
Since \(\sin 30° = \frac{1}{2}\):
\[ 1.5 = \frac{\sin I_2}{\frac{1}{2}} \]
This gives:
\[ \sin I_2 = 1.5 \times \frac{1}{2} = 0.75 \]

### Step 6: Calculate the Angle I2
To find I2:
\[ I_2 = \sin^{-1}(0.75) \]
Calculating this gives:
\[ I_2 \approx 48.6° \]

### Step 7: Calculate the Angle of Deviation (Δ)
The angle of deviation (Δ) is given by:
\[ \Delta = I_1 + I_2 - A \]
Substituting the values:
\[ \Delta = 0° + 48.6° - 30° \]
Thus:
\[ \Delta = 18.6° \]

### Final Answer
The angle of deviation (Δ) between the incident ray and the ray that leaves the prism is approximately **18.6°**.

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