Light is incident normally on a diffraction grating through which the first order diffraction is seen at `32^@`. The second order diffraction will be seen at

Options

A. at`80^@`
B. at`64^@`
C. at`48^@`
D. there is no ssecond order diffraction

Explanation

To solve the problem, we need to use the diffraction grating formula, which relates the angle of diffraction to the wavelength of light and the grating spacing. The formula is given by:

\[ d \sin \theta_n = n \lambda \]

where:
- \( d \) is the grating spacing (distance between adjacent slits),
- \( \theta_n \) is the angle of diffraction for the nth order,
- \( n \) is the order of the diffraction,
- \( \lambda \) is the wavelength of the light.

### Step 1: Understand the first order diffraction condition
Given that the first order diffraction (\( n = 1 \)) occurs at an angle of \( \theta_1 = 32^\circ \), we can write:

\[ d \sin(32^\circ) = 1 \cdot \lambda \]

### Step 2: Calculate \( \sin(32^\circ) \)
Using a calculator or sine table, we find:

\[ \sin(32^\circ) \approx 0.5299 \]

### Step 3: Substitute into the equation
Now substituting this value into the equation, we have:

\[ d \cdot 0.5299 = \lambda \]

This gives us the relationship between the grating spacing \( d \) and the wavelength \( \lambda \).

### Step 4: Set up the second order diffraction condition
For the second order diffraction (\( n = 2 \)), we can use the same formula:

\[ d \sin(\theta_2) = 2 \lambda \]

### Step 5: Substitute \( \lambda \) from the first order
From the first order diffraction, we know that \( \lambda = d \cdot 0.5299 \). Substituting this into the second order equation gives:

\[ d \sin(\theta_2) = 2(d \cdot 0.5299) \]

### Step 6: Simplify the equation
Canceling \( d \) from both sides (assuming \( d \neq 0 \)) gives:

\[ \sin(\theta_2) = 2 \cdot 0.5299 = 1.0598 \]

### Step 7: Analyze the result
Since the sine of an angle cannot exceed 1, we conclude that:

\[ \sin(\theta_2) > 1 \]

This indicates that second order diffraction is not possible under these conditions.

### Final Answer
Thus, the second order diffraction will not be seen.