Light is incident from glass `(mu = 1.50)` to water `(mu = 1.33)` find the range of the angle of deviation for which there are two angles of incidence.

Explanation

To solve the problem of finding the range of the angle of deviation for light incident from glass (μ = 1.50) to water (μ = 1.33), we can follow these steps:

### Step 1: Identify the refractive indices
We have:
- Refractive index of glass (μ₁) = 1.50 (denser medium)
- Refractive index of water (μ₂) = 1.33 (rarer medium)

### Step 2: Determine the critical angle
The critical angle (θ_C) can be found using Snell's law. The formula for the critical angle when light travels from a denser medium to a rarer medium is given by:

\[ \sin(\theta_C) = \frac{\mu_2}{\mu_1} \]

Substituting the values:

\[ \sin(\theta_C) = \frac{1.33}{1.50} = \frac{4}{3} \cdot \frac{1}{1.5} = \frac{8}{9} \]

### Step 3: Calculate the critical angle
Now, we calculate θ_C using the inverse sine function:

\[ \theta_C = \sin^{-1}\left(\frac{8}{9}\right) \]

Calculating this gives:

\[ \theta_C \approx 62.73^\circ \]

### Step 4: Calculate the angle of deviation at critical angle
The angle of deviation (Δ) is given by:

\[ \Delta = 90^\circ - \theta_C \]

Substituting the value of θ_C:

\[ \Delta = 90^\circ - 62.73^\circ \approx 27.27^\circ \]

### Step 5: Determine the range of the angle of deviation
For the range of angles of deviation where there are two angles of incidence, we consider:
- At normal incidence (θ = 0), the deviation is 0°.
- At the critical angle (θ_C), the deviation is approximately 27.27°.

Thus, the range of the angle of deviation for which there are two angles of incidence is:

\[ \Delta \text{ ranges from } 0^\circ \text{ to } 27.27^\circ \]

### Final Answer
The range of the angle of deviation for which there are two angles of incidence is:

\[ \Delta: 0^\circ \text{ to } 27.27^\circ \]

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