If a ray of light is incident from rarer medium at an angle 45° on the surface which separates two medium having refractive indices `1` and `sqrt(2)` for rarer and denser medium, then angle of deviation of refractive ray with incident ray is

Options

A. 45°
B. 15°
C. 30°
D. 75°

Answer

To solve the problem, we need to find the angle of deviation of a refracted ray when a ray of light is incident at an angle of 45° from a rarer medium (refractive index = 1) to a denser medium (refractive index = √2).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Angle of incidence (i) = 45°
- Refractive index of rarer medium (n1) = 1
- Refractive index of denser medium (n2) = √2

2. **Apply Snell's Law:**
Snell's Law states that:
\[ n_1 \sin(i) = n_2 \sin(r) \]
where \( r \) is the angle of refraction.

3. **Substitute the Known Values:**
\[ 1 \cdot \sin(45°) = \sqrt{2} \cdot \sin(r) \]
Since \( \sin(45°) = \frac{1}{\sqrt{2}} \), we can substitute this in:
\[ 1 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin(r) \]

4. **Solve for \( \sin(r) \):**
Rearranging gives:
\[ \sin(r) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \]

5. **Find the Angle of Refraction \( r \):**
The angle whose sine is \( \frac{1}{2} \) is:
\[ r = 30° \]

6. **Calculate the Angle of Deviation:**
The angle of deviation \( D \) is given by:
\[ D = i - r \]
Substituting the values:
\[ D = 45° - 30° = 15° \]

### Final Answer:
The angle of deviation of the refracted ray with respect to the incident ray is **15°**.

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