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 find the wavelength of light used in the diffraction experiment, we use the formula for the width of the central maximum in a single slit diffraction pattern:
\[ w = \frac{2 \lambda L}{a} \]
where \( w \) is the linear width of the central maximum, \( \lambda \) is the wavelength of light, \( L \) is the distance from the slit to the screen, and \( a \) is the width of the slit.
Given:

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

Substitute the known values into the formula:
\[ 5 \times 10^{-3} = \frac{2 \lambda \times 0.5}{0.1 \times 10^{-3}} \]
Simplify and solve for \( \lambda \):
\[ 5 \times 10^{-3} = \frac{1 \times \lambda}{0.1 \times 10^{-3}} \]
\[ \lambda = \frac{5 \times 10^{-3} \times 0.1 \times 10^{-3}}{1} \]
\[ \lambda = 5 \times 10^{-7} \ \text{m} \]
Thus, the wavelength of light used is \( 5 \times 10^{-7} \ \text{m} \).