A ray of light travelling in air is incident at ` 45^(@)` on a medium of refractive index ` sqrt(2)`. The angle of refractive in the medium is

Options

A. A. ` 45^(@)`
B. B.`60^(@)`
C. C.`30^(@)`
D. D.`70^(@)`

Explanation

To solve the problem of finding the angle of refraction when a ray of light travels from air into a medium with a refractive index of \(\sqrt{2}\), we can use Snell's Law. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a ray of light incident at an angle of \(45^\circ\) from air (where the refractive index \(n_1 = 1\)) into a medium with a refractive index \(n_2 = \sqrt{2}\). We need to find the angle of refraction \(r\) in the medium.

### Step 2: Apply Snell's Law
Snell's Law states that:
\[ n_1 \sin(i) = n_2 \sin(r) \]
Where:
- \(n_1\) = refractive index of the first medium (air) = 1
- \(i\) = angle of incidence = \(45^\circ\)
- \(n_2\) = refractive index of the second medium = \(\sqrt{2}\)
- \(r\) = angle of refraction

### Step 3: Substitute the Known Values
Substituting the known values into Snell's Law:
\[ 1 \cdot \sin(45^\circ) = \sqrt{2} \cdot \sin(r) \]

### Step 4: Calculate \(\sin(45^\circ)\)
We know that:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
So we can rewrite the equation as:
\[ 1 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin(r) \]

### Step 5: Rearrange the Equation
Rearranging gives:
\[ \sin(r) = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2} \]

### Step 6: Find the Angle \(r\)
Now we need to find the angle \(r\) such that:
\[ \sin(r) = \frac{1}{2} \]
The angle \(r\) that satisfies this equation is:
\[ r = 30^\circ \]

### Conclusion
The angle of refraction in the medium is \(30^\circ\).

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