Find the maximum angle that can be made in glass medium `(mu=1.5)` if a light ray is refreacted from glass to vacuum.

Options

A. `sin^(-1)4//3`
B. `sin^(-1)2//3`
C. `sin^(-1)1//3`
D. `sin^(-1)5//3`

Explanation

To find the maximum angle that can be made in a glass medium (with a refractive index \( \mu = 1.5 \)) when a light ray is refracted from glass to vacuum, we can follow these steps:

### Step 1: Understand the situation
We are dealing with the refraction of light from glass (with refractive index \( \mu = 1.5 \)) to vacuum (with refractive index \( n = 1 \)). We want to find the maximum angle of incidence in the glass medium.

### Step 2: Apply Snell's Law
Snell's Law states that:
\[ \mu_1 \sin i = \mu_2 \sin r \]
where:
- \( \mu_1 \) is the refractive index of the first medium (glass, \( \mu_1 = 1.5 \))
- \( \mu_2 \) is the refractive index of the second medium (vacuum, \( \mu_2 = 1 \))
- \( i \) is the angle of incidence in the glass
- \( r \) is the angle of refraction in the vacuum

### Step 3: Determine the maximum angle of refraction
For the maximum angle of incidence, the angle of refraction \( r \) will be \( 90^\circ \). This is because at this angle, the light ray will just graze the boundary between the two media.

### Step 4: Substitute values into Snell's Law
Substituting the known values into Snell's Law gives:
\[ 1.5 \sin i = 1 \sin 90^\circ \]
Since \( \sin 90^\circ = 1 \), we can simplify this to:
\[ 1.5 \sin i = 1 \]

### Step 5: Solve for \( \sin i \)
Rearranging the equation:
\[ \sin i = \frac{1}{1.5} = \frac{2}{3} \]

### Step 6: Find the angle \( i \)
To find the angle \( i \), we take the inverse sine:
\[ i = \sin^{-1} \left( \frac{2}{3} \right) \]

### Final Answer
Thus, the maximum angle that can be made in the glass medium is:
\[ i = \sin^{-1} \left( \frac{2}{3} \right) \]