Light from a luminious point on the lower face of a rectangular glass slab , 2 cm thick, strikes the upper face and the totally reflected rays outline a circle of 3.2 cm radius on the lower face. Find the refractive index of the glass.

Explanation

To solve the problem, we need to find the refractive index of the glass slab given the thickness of the slab and the radius of the circle outlined by the totally reflected rays on the lower face.

### Step-by-Step Solution:

1. **Understand the Geometry**:
- We have a rectangular glass slab with a thickness (t) of 2 cm.
- A luminous point is located on the lower face of the slab.
- The rays of light strike the upper face and undergo total internal reflection, outlining a circle of radius (r) = 3.2 cm on the lower face.

2. **Identify the Angles**:
- Let’s denote the angle of incidence at the upper face as \( C \).
- The angle of refraction at the upper face is \( 90^\circ \) since the rays are totally internally reflected.

3. **Apply Snell's Law**:
- According to Snell's Law:
\[ \mu_1 \sin C = \mu_2 \sin 90^\circ \]
- Here, \( \mu_2 \) (the refractive index of air) is 1, and \( \sin 90^\circ = 1 \).
- Therefore, we can simplify this to:
\[ \mu_1 \sin C = 1 \]
- Rearranging gives:
\[ \mu_1 = \frac{1}{\sin C} \]

4. **Determine the Geometry of the Circle**:
- The radius of the circle (r) formed on the lower face is 3.2 cm.
- The distance from the luminous point to the upper face is the thickness of the slab (t = 2 cm).
- We can visualize a right triangle where:
- The opposite side is the radius (r = 3.2 cm).
- The adjacent side is the thickness of the slab (t = 2 cm).
- Using the Pythagorean theorem, we can find the hypotenuse (h):
\[ h = \sqrt{r^2 + t^2} = \sqrt{(3.2)^2 + (2)^2} \]
\[ h = \sqrt{10.24 + 4} = \sqrt{14.24} \approx 3.77 \text{ cm} \]

5. **Calculate \( \sin C \)**:
- From the triangle, we can find \( \sin C \):
\[ \sin C = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{r}{h} = \frac{3.2}{3.77} \approx 0.847 \]

6. **Find the Refractive Index**:
- Now, substituting \( \sin C \) back into the equation for \( \mu_1 \):
\[ \mu_1 = \frac{1}{\sin C} = \frac{1}{0.847} \approx 1.18 \]

### Final Answer:
The refractive index of the glass slab is approximately **1.18**.