The `I^(st)` diffraction minimum due to single slit diffraction is `theta`, for a light of wave length `5000 Å` . If the width of the slit si `1xx10^(-4) cm` then the value of `theta`

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `15^@`

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the given values into standard units
We are given the wavelength of light, \( \lambda = 5000 \, \text{Å} \), and the width of the slit, \( d = 1 \times 10^{-4} \, \text{cm} \).

1. Convert wavelength from angstroms to meters:
\[ \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]

2. Convert the width of the slit from centimeters to meters:
\[ d = 1 \times 10^{-4} \, \text{cm} = 1 \times 10^{-6} \, \text{m} \]

### Step 2: Use the formula for the first diffraction minimum
For single slit diffraction, the condition for the first minimum is given by:
\[ d \sin \theta = n \lambda \]
where \( n \) is the order of the minimum. For the first minimum, \( n = 1 \).

Substituting the values we have:
\[ 1 \times 10^{-6} \sin \theta = 1 \times (5 \times 10^{-7}) \]

### Step 3: Solve for \( \sin \theta \)
Rearranging the equation to find \( \sin \theta \):
\[ \sin \theta = \frac{5 \times 10^{-7}}{1 \times 10^{-6}} = 0.5 \]

### Step 4: Calculate \( \theta \)
Now, we need to find the angle \( \theta \) such that \( \sin \theta = 0.5 \).

From trigonometric values, we know:
\[ \sin 30^\circ = 0.5 \]
Thus,
\[ \theta = 30^\circ \]

### Final Answer
The value of \( \theta \) is \( 30^\circ \).

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