In Young's double-slit experiment, interference pattern is observed at a distance of 1m. If light used has a frequency of 6 × 10 14 Hz and slit separation distance is equal to 0.05 mm,, calculate the angular position of `8^(th)` maximum.

Answer

To solve the problem of finding the angular position of the 8th maximum in Young's double-slit experiment, we can follow these steps:

### Step 1: Identify the given parameters
- Frequency of light, \( f = 6 \times 10^{14} \) Hz
- Slit separation distance, \( d = 0.05 \) mm = \( 0.05 \times 10^{-3} \) m = \( 5 \times 10^{-5} \) m
- Distance from slits to screen, \( D = 1 \) m
- Order of maximum, \( n = 8 \)

### Step 2: Calculate the wavelength \( \lambda \)
Using the formula for the speed of light:
\[ c = f \lambda \]
where \( c \) (speed of light) is approximately \( 3 \times 10^8 \) m/s. Rearranging gives:
\[ \lambda = \frac{c}{f} \]
Substituting the values:
\[ \lambda = \frac{3 \times 10^8 \, \text{m/s}}{6 \times 10^{14} \, \text{Hz}} = 0.5 \times 10^{-6} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]

### Step 3: Calculate the position \( y_n \) of the 8th maximum
The position of the nth maximum is given by:
\[ y_n = \frac{n \lambda D}{d} \]
Substituting \( n = 8 \), \( \lambda = 5 \times 10^{-7} \, \text{m} \), \( D = 1 \, \text{m} \), and \( d = 5 \times 10^{-5} \, \text{m} \):
\[ y_8 = \frac{8 \times (5 \times 10^{-7}) \times 1}{5 \times 10^{-5}} = \frac{40 \times 10^{-7}}{5 \times 10^{-5}} = \frac{40}{5} \times 10^{-2} = 8 \times 10^{-2} \, \text{m} \]

### Step 4: Calculate the angular position \( \theta \)
The angular position \( \theta \) can be calculated using:
\[ \tan \theta = \frac{y_n}{D} \]
For small angles, \( \tan \theta \approx \theta \) (in radians):
\[ \theta \approx \frac{y_n}{D} = \frac{8 \times 10^{-2}}{1} = 8 \times 10^{-2} \, \text{radians} \]

### Final Answer
The angular position of the 8th maximum is:
\[ \theta \approx 8 \times 10^{-2} \, \text{radians} \]

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