Word ‘Newton’s printed on a paper and is placed on a horizontal surface below a cubical glass. The minimum value of refractive index of a cubical glass for which letters are not visible from any vertical faces, of the glass, is (Critical angle =45°)

Options

A. `sqrt(3)`
B. 0.5
C. 1
D. `sqrt(2)`

Explanation

To solve the problem, we need to determine the minimum value of the refractive index of a cubical glass such that the word "Newton" printed on a paper beneath it is not visible from any of the vertical faces of the glass. Given that the critical angle (θc) is 45°, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Critical Angle**:
The critical angle is the angle of incidence above which total internal reflection occurs. For a given refractive index \( n \), the critical angle \( \theta_c \) can be expressed using Snell's Law:
\[ \sin(\theta_c) = \frac{n_2}{n_1} \]
where \( n_1 \) is the refractive index of the denser medium (glass) and \( n_2 \) is the refractive index of the rarer medium (air, which is approximately 1).

2. **Setting Up the Equation**:
Since \( \theta_c = 45^\circ \), we have:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.707 \]
Therefore, substituting into the equation:
\[ \frac{1}{\sqrt{2}} = \frac{1}{n_1} \]
Rearranging gives:
\[ n_1 = \sqrt{2} \]

3. **Interpreting the Result**:
The refractive index \( n_1 \) of the glass must be at least \( \sqrt{2} \) (approximately 1.414) for the light rays coming from the letters to undergo total internal reflection. This means that no light can escape through the vertical faces of the glass, making the letters invisible.

4. **Conclusion**:
The minimum value of the refractive index of the cubical glass for which the letters "Newton" are not visible from any vertical face is:
\[ n_1 = \sqrt{2} \]

### Final Answer:
The minimum value of the refractive index of the cubical glass is \( \sqrt{2} \).

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