Question

In a diffraction experiment, the fringe width $ \beta $ is 0.3 mm, the distance from the slit to the screen $ D $ is 5 cm, and the slit width $ d $ is 3 mm. What is the wavelength $ \lambda $?

• A. 500 nm
• B. 600 nm
• C. 400 nm
• D. 300 nm

Hint:

Use the diffraction formula $ \beta = \frac{\lambda D}{d} $ to find the unknown variable. Always convert all units to meters before substituting into the formula.

Answer 1

The Correct Option is B

Step 1: Recall the formula for fringe width in diffraction
The fringe width $ \beta $ in single-slit diffraction is given by:
\[ \beta = \frac{\lambda D}{d} \] where:
$ \beta $ = fringe width
$ \lambda $ = wavelength of light
$ D $ = distance from the slit to the screen
$ d $ = slit width
Step 2: Convert all units to meters
Given:
$ \beta = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} $
$ D = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} $
$ d = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} $
Step 3: Rearranging the formula to solve for \(\lambda\)
\[ \lambda = \frac{\beta \cdot d}{D} \] Substitute the known values:
\[ \lambda = \frac{(0.3 \times 10^{-3}) \times (3 \times 10^{-3})}{5 \times 10^{-2}} = \frac{0.9 \times 10^{-6}}{5 \times 10^{-2}} = 0.18 \times 10^{-4} \text{ m} = 600 \times 10^{-9} \text{ m} \] So,
\[ \lambda = 600 \text{ nm} \] Step 4: Conclusion
The wavelength of the light used is:
\[ (B) 600 nm \]