A ray of light is incident on a glass slab of refractive index 1.5, making an angle of `40^(@)` with the surface. The angle of refraction in glass is

Options

A. `40^(@)`
B. `30^(@) 42'`
C. `50^(@)`
D. `60^(@)`

Explanation

To find the angle of refraction in the glass slab when a ray of light is incident at an angle of \(40^\circ\) with the surface, we can follow these steps:

### Step 1: Understand the Given Information
- The refractive index of the glass slab, \(n = 1.5\).
- The angle of incidence with respect to the surface is \(40^\circ\).

### Step 2: Determine the Angle of Incidence with Respect to the Normal
The angle of incidence with respect to the normal can be calculated as:
\[ i = 90^\circ - 40^\circ = 50^\circ \]

### Step 3: Apply Snell's Law
Snell's Law states that:
\[ n_1 \sin(i) = n_2 \sin(r) \]
Where:
- \(n_1\) is the refractive index of the first medium (air, approximately \(1\)).
- \(n_2\) is the refractive index of the second medium (glass, \(1.5\)).
- \(i\) is the angle of incidence (which we found to be \(50^\circ\)).
- \(r\) is the angle of refraction.

Substituting the known values into Snell's Law:
\[ 1 \cdot \sin(50^\circ) = 1.5 \cdot \sin(r) \]

### Step 4: Solve for \(\sin(r)\)
Rearranging the equation gives:
\[ \sin(r) = \frac{\sin(50^\circ)}{1.5} \]

### Step 5: Calculate \(\sin(50^\circ)\)
Using a calculator, we find:
\[ \sin(50^\circ) \approx 0.7660 \]

### Step 6: Substitute and Calculate \(\sin(r)\)
Now substituting the value of \(\sin(50^\circ)\):
\[ \sin(r) = \frac{0.7660}{1.5} \approx 0.5107 \]

### Step 7: Find the Angle of Refraction \(r\)
Now, we take the inverse sine to find \(r\):
\[ r = \sin^{-1}(0.5107) \approx 30.42^\circ \]

### Final Answer
The angle of refraction in the glass is approximately \(30.42^\circ\).

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