Question

In Young's double slit experiment, carried out with light of wavelength 5000Å, the distance between the slits is 0.3 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0 cm. The value of x for third maxima is ............. mm.

Answer 1

Correct Answer: 10

\[ \beta = \frac{\lambda D}{d} = \frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}} = 10 \times 10^{-3} \, \text{m} \]

For 3rd maxima:

\[ y_3 = 3\beta = 10 \times 10^{-3} \, \text{m} = 10 \, \text{mm} \]

Answer 2

Step 1: Given data
Wavelength, \( \lambda = 5000 \, \text{Å} = 5 \times 10^{-7} \, \text{m} \)
Distance between the slits, \( d = 0.3 \, \text{mm} = 3 \times 10^{-4} \, \text{m} \)
Distance between the screen and the slits, \( D = 200 \, \text{cm} = 2 \, \text{m} \).

Step 2: Formula for position of bright fringes (maxima)
The position of the \( n^{th} \) bright fringe (maxima) from the central maximum is given by: \[ x_n = n \frac{\lambda D}{d} \] For the **third maxima** \( (n = 3) \):
\[ x_3 = 3 \frac{\lambda D}{d}. \]

Step 3: Substitute the values
\[ x_3 = 3 \times \frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}} = 3 \times \frac{1 \times 10^{-6}}{3 \times 10^{-4}} = 3 \times 3.33 \times 10^{-3} = 1 \times 10^{-2} \, \text{m}. \]

Step 4: Convert to millimeters
\[ 1 \times 10^{-2} \, \text{m} = 10 \, \text{mm}. \]

Final answer

10 mm