When a beam of light goes from denser medium (`mu_d`) to rarer medium (`mu_r`), then it is generally observed that magnitude of angle of incidence is half that of angle of refraction. Then magnitude of incident angle will be-(here `mu=mu_d //mu_r`)

Options

A. `2sin^(-1) (mu/2)`
B. `2 cos^(-1)mu`
C. `cos^(-1)(mu/2)`
D. `2cos^(-1)(mu/2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between angles of incidence and refraction
Given that the angle of incidence \( I \) is half of the angle of refraction \( R \), we can express this relationship as:
\[ I = \frac{R}{2} \]
or equivalently,
\[ R = 2I \]

### Step 2: Apply Snell's Law
Snell's Law states that:
\[ \frac{\sin I}{\sin R} = \frac{\mu_r}{\mu_d} \]
Given that \( \mu = \frac{\mu_d}{\mu_r} \), we can rewrite \( \frac{\mu_r}{\mu_d} \) as \( \frac{1}{\mu} \). Thus, we have:
\[ \frac{\sin I}{\sin R} = \frac{1}{\mu} \]

### Step 3: Substitute for \( R \)
Substituting \( R = 2I \) into Snell's Law gives us:
\[ \frac{\sin I}{\sin(2I)} = \frac{1}{\mu} \]

### Step 4: Use the double angle formula
Using the double angle formula for sine, we know that:
\[ \sin(2I) = 2 \sin I \cos I \]
Substituting this into our equation from Step 3, we get:
\[ \frac{\sin I}{2 \sin I \cos I} = \frac{1}{\mu} \]

### Step 5: Simplify the equation
Cancelling \( \sin I \) from both sides (assuming \( \sin I \neq 0 \)):
\[ \frac{1}{2 \cos I} = \frac{1}{\mu} \]
Cross-multiplying gives:
\[ \mu = 2 \cos I \]

### Step 6: Solve for \( I \)
Rearranging the equation gives:
\[ \cos I = \frac{\mu}{2} \]
To find \( I \), we take the inverse cosine:
\[ I = \cos^{-1}\left(\frac{\mu}{2}\right) \]

### Final Result
Thus, the magnitude of the incident angle \( I \) is:
\[ I = \cos^{-1}\left(\frac{\mu}{2}\right) \]