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 the refracted ray with respect to the incident ray when a ray of light passes from a rarer medium to a denser medium.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Angle of incidence (i) = 45°
- Refractive index of rarer medium (μ1) = 1
- Refractive index of denser medium (μ2) = √2

2. **Apply Snell's Law:**
Snell's Law states that:
\[ \mu_1 \sin(i) = \mu_2 \sin(r) \]
where \( r \) is the angle of refraction.

3. **Substitute the Known Values:**
\[ 1 \cdot \sin(45°) = \sqrt{2} \cdot \sin(r) \]
We know that \( \sin(45°) = \frac{\sqrt{2}}{2} \), so:
\[ \frac{\sqrt{2}}{2} = \sqrt{2} \cdot \sin(r) \]

4. **Simplify the Equation:**
Dividing both sides by \( \sqrt{2} \):
\[ \frac{1}{2} = \sin(r) \]

5. **Find the Angle of Refraction (r):**
To find \( r \), we take the inverse sine:
\[ r = \sin^{-1}\left(\frac{1}{2}\right) \]
This gives us:
\[ r = 30° \]

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

7. **Final Answer:**
The angle of deviation of the refracted ray with respect to the incident ray is:
\[ \Delta = 15° \]