Lights of wavelength 550 nm falls normally on a slit of width k`22.0 xx 10^(-5) cm`. The angular position of the central maximum will be (in radian) :

Options

A. `pi/6`
B. `(pi)/(4)`
C. `(pi)/(8)`
D. `(pi)/(12)`

Answer

To solve the problem of finding the angular position of the central maximum when light of wavelength 550 nm falls normally on a slit of width \( 2.0 \times 10^{-5} \) cm, we can follow these steps:

### Step 1: Convert the units of wavelength and slit width
- The wavelength \( \lambda \) is given as 550 nm. We need to convert this to meters:
\[ \lambda = 550 \, \text{nm} = 550 \times 10^{-9} \, \text{m} \]
- The slit width \( k \) is given as \( 2.0 \times 10^{-5} \) cm. We need to convert this to meters:
\[ k = 2.0 \times 10^{-5} \, \text{cm} = 2.0 \times 10^{-5} \times 10^{-2} \, \text{m} = 2.0 \times 10^{-7} \, \text{m} \]

### Step 2: Use the formula for the angular position of the central maximum
The formula for the sine of the angle \( \theta \) for the central maximum in a single-slit diffraction pattern is given by:
\[ \sin \theta = \frac{\lambda}{k} \]

### Step 3: Substitute the values into the formula
Now we can substitute the values of \( \lambda \) and \( k \) into the formula:
\[ \sin \theta = \frac{550 \times 10^{-9}}{2.0 \times 10^{-7}} \]

### Step 4: Calculate \( \sin \theta \)
Calculating the above expression:
\[ \sin \theta = \frac{550 \times 10^{-9}}{2.0 \times 10^{-7}} = \frac{550}{2.0} \times 10^{-2} = 275 \times 10^{-2} = 0.275 \]

### Step 5: Find the angle \( \theta \)
To find the angle \( \theta \), we take the inverse sine:
\[ \theta = \arcsin(0.275) \]

Using a calculator, we find:
\[ \theta \approx 0.281 \, \text{radians} \]

### Step 6: Final answer
Thus, the angular position of the central maximum is approximately:
\[ \theta \approx 0.281 \, \text{radians} \]