Question

A screen is placed 50 cm from a single slit which is illuminated with light of wavelength 6000Å. If the distance between the first and third minima in the diffraction pattern is 3.0mm. The width of the slit is:

• A. 1×10^-4 m
• B. 2×10^-4 m
• C. 0.5×10^-4 m
• D. 4×10^-4 m

Answer 1

Answer: (B)

2×10^-4 m

Answer 2

The position of nth minima in the diffreaction pattern is

$x_{n} = \frac{nD\lambda}{d}$

$\therefore x_{3} - x_{1} = (3 - 1)\frac{D\lambda}{d} = \frac{2D\lambda}{d}$

Or $d = \frac{2D\lambda}{x_{3} - x_{1}} = \frac{2 \times 0.50 \times 6000 \times 10^{- 10}}{3 \times 10^{- 3}} = 2 \times 10^{- 4}m$