If refractive index of glass is 1.50 and of water is 1.33, then criticle angle is

Options

A. `sin^(-1)(8/9)`
B. `sin^(-1)(2/3)`
C. `cos^(-1)(8/9)`
D. None of the above

Explanation

To find the critical angle when light travels from glass to water, we can use the formula for the critical angle, which is given by:

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

where:
- \(\theta_c\) is the critical angle,
- \(\mu_1\) is the refractive index of the rarer medium (water in this case),
- \(\mu_2\) is the refractive index of the denser medium (glass in this case).

### Step-by-Step Solution:

1. **Identify the refractive indices:**
- Refractive index of glass (\(\mu_2\)) = 1.50
- Refractive index of water (\(\mu_1\)) = 1.33

2. **Apply the formula for the critical angle:**
\[ \sin(\theta_c) = \frac{\mu_1}{\mu_2} = \frac{1.33}{1.50} \]

3. **Calculate the value:**
\[ \sin(\theta_c) = \frac{1.33}{1.50} \approx 0.8867 \]

4. **Find the critical angle using the inverse sine function:**
\[ \theta_c = \sin^{-1}(0.8867) \]

5. **Calculate \(\theta_c\):**
Using a calculator, we find:
\[ \theta_c \approx 61.0^\circ \]

### Final Answer:
The critical angle \(\theta_c\) is approximately \(61.0^\circ\).

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