A beam of monochromatic light is incident at an angle of `55^(@)` on one face of an equilateral prism . The angle of deviation is `delta` and the angle of emergence is `46^(@)` . If the angle of minimum deviation `(delta_(m))` for the same prism is `45^(@)` , then

Options

A. `delta = delta_(m)`
B. `delta gt delta_(m)`
C. `delta lt delta_(m)`
D. `delta = (2)/(3) delta_(m)`

Explanation

To solve the problem step by step, we will use the relationship between the angles of incidence, emergence, the angle of the prism, and the angle of deviation.

### Step 1: Identify the given values
- Angle of incidence (I) = 55°
- Angle of emergence (E) = 46°
- Angle of the prism (A) = 60° (since it is an equilateral prism)
- Angle of minimum deviation (δm) = 45°

### Step 2: Use the formula relating the angles
The relationship between the angles is given by:
\[ I + E = A + \delta \]

### Step 3: Substitute the known values into the formula
Substituting the known values into the equation:
\[ 55° + 46° = 60° + \delta \]

### Step 4: Simplify the equation
Now, we simplify the left side:
\[ 101° = 60° + \delta \]

### Step 5: Solve for the angle of deviation (δ)
To find δ, we rearrange the equation:
\[ \delta = 101° - 60° \]
\[ \delta = 41° \]

### Step 6: Compare δ with δm
Now, we compare the calculated angle of deviation (δ) with the angle of minimum deviation (δm):
- δ = 41°
- δm = 45°

Since δ (41°) is less than δm (45°), we conclude that:
\[ \delta < \delta_m \]

### Final Conclusion
The angle of deviation (δ) is less than the angle of minimum deviation (δm).

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