A single slit is illuminated by light of wavelength 6000 Ã…. The slit width is 0.1 cm and the screen is placed 1 m away. The angular position of `10^(th)` minimum in radian is

Options

A. `2 xx 10^(-4)`
B. `6 xx 10^(-3)`
C. `3 xx 10^(-3)`
D. `1 xx 10^(-4)`

Explanation

To find the angular position of the 10th minimum in a single slit diffraction pattern, we can use the formula for the position of the minima:

\[ \sin \theta_n = \frac{n \lambda}{d} \]

Where:
- \( n \) is the order of the minimum (in this case, \( n = 10 \)),
- \( \lambda \) is the wavelength of the light,
- \( d \) is the width of the slit.

### Step 1: Convert the given values to appropriate units
- Wavelength \( \lambda = 6000 \) Å = \( 6000 \times 10^{-10} \) m = \( 6 \times 10^{-7} \) m
- Slit width \( d = 0.1 \) cm = \( 0.1 \times 10^{-2} \) m = \( 1 \times 10^{-3} \) m

### Step 2: Substitute the values into the formula
Using the formula for the angular position of the minima, we substitute \( n = 10 \), \( \lambda = 6 \times 10^{-7} \) m, and \( d = 1 \times 10^{-3} \) m:

\[ \sin \theta_{10} = \frac{10 \times 6 \times 10^{-7}}{1 \times 10^{-3}} \]

### Step 3: Calculate the value
Calculating the right side:

\[ \sin \theta_{10} = \frac{60 \times 10^{-7}}{1 \times 10^{-3}} = 60 \times 10^{-4} = 6 \times 10^{-3} \]

### Step 4: Find the angle in radians
Since for small angles, \( \sin \theta \approx \theta \) (in radians), we can approximate:

\[ \theta_{10} \approx 6 \times 10^{-3} \text{ radians} \]

### Final Answer
The angular position of the 10th minimum in radians is:

\[ \theta_{10} = 6 \times 10^{-3} \text{ radians} \]

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