Question

A laser shines on a pair of vertical slits. The horizontal distance L1 between the laser and the ruler is 12.3 m. The distance L2 between the laser and the slits is 0.511 m. The distance d is 0.440 mm. A laser is located a distance L1 from a ruler and a distance L2 from a barrier containing two, narrow slits. The slit separation is d. An interference intensity pattern is shown at the position of the ruler. The intensity pattern has a central maximum at 0 centimeters. The first dark fringe is at 1.6 centimeters. The first bright fringe is at about 3 centimeters. The second dark fringe is at about 4.8 centimeters. The second bright fringe is at about 6.4 centimeters. The third dark fringe is at 8 centimeters. The illustration is not to scale. Note that the ruler measures in centimeters. Calculate the wavelength ?

Answer 1

Answer:

Wavelength = 1.12 μm

Explanation:

Fringe width = Distance between consecutive maximums or consecutive minimums

Using the first bright fringe to calculate the wavelength

The first bright fringe, $\beta = 3 cm = 3/100 = 0.03 m$

Distance between the laser and the ruler, $L_{1} = 12.3 m$

Distance between the laser and the slits, $L_{2} = 0.511 m$

Distance between the ruler and the slits, $D = L_{1} - L_{2}$

$D = 12.3 - 0.511$

$D = 11.789 m$

distance, d = 0.440 mm = 0.00044 m

The relationship between the fringe width and the wavelength is given by:

The fringe width, $\beta = \frac{\lambda D}{d}$

$0.03 = \frac{\lambda 11.789}{0.00044} \\0.03 * 0.00044 = \lambda 11.789\\\lambda = \frac{0.03*0.00044}{11.789} \\\lambda = 1.12 * 10^{-6} m \\\lambda = 1.12 \mu m$