Question

Using light from a monochromatic source to study diffraction in a single slit of width 0.1 mm, the linear width of central maxima is measured to be 5 mm on a screen held 50 cm away. The wavelength of light used is _______.
Fill in the blank with the correct answer from the options given below

• A. \(2.5 \times 10^{-7} \ \text{m}\)
• B. \(4 \times 10^{-7} \ \text{m}\)
• C. \(5 \times 10^{-7} \ \text{m}\)
• D. \(7.5 \times 10^{-7} \ \text{m}\)

Answer 1

The Correct Option is C

To determine the wavelength of light used in the diffraction experiment, the formula for the central maximum width in a single-slit diffraction pattern is applied:
\[ w = \frac{2 \lambda L}{a} \]
Here, \( w \) represents the linear width of the central maximum, \( \lambda \) is the light's wavelength, \( L \) is the slit-to-screen distance, and \( a \) is the slit width.
Provided data:

• \( w = 5 \ \text{mm} = 5 \times 10^{-3} \ \text{m} \)
• \( L = 50 \ \text{cm} = 0.5 \ \text{m} \)
• \( a = 0.1 \ \text{mm} = 0.1 \times 10^{-3} \ \text{m} \)

Substituting these values into the formula yields:
\[ 5 \times 10^{-3} = \frac{2 \lambda \times 0.5}{0.1 \times 10^{-3}} \]
Simplification and solving for \( \lambda \) proceeds as follows:
\[ 5 \times 10^{-3} = \frac{1 \times \lambda}{0.1 \times 10^{-3}} \]
\[ \lambda = \frac{5 \times 10^{-3} \times 0.1 \times 10^{-3}}{1} \]
\[ \lambda = 5 \times 10^{-7} \ \text{m} \]
Therefore, the wavelength of the light employed is \( 5 \times 10^{-7} \ \text{m} \).