Question

Determine the distance that the third bright fringe would lie from the central bisector in a single slit diffraction pattern generated with 542 nm light incident on a 1.2 x 10-4 m slit falling onto a screen 68cm away.

Answer 1

Answer on Question #66031, Physics / Other

Determine the distance that the third bright fringe would lie from the central bisector in a single slit diffraction pattern generated with $542\mathrm{nm}$ light incident on a $1.2\times 10^{-4}\mathrm{m}$ slit falling onto a screen $68\mathrm{cm}$ away.

Solution:

The general condition for a minimum for a single slit is:

$$m \lambda = a \sin \theta$$

where $m = 1,2,3,4$ and so on

- $a$ is the width of the slit,

- $\theta$ is the angle of incidence at which the minimum intensity occurs, and

- $\lambda$ is the wavelength of the light

The distance of the bright fringe on screen from the central bisector is:

$$y \approx \frac {m \lambda D}{a}$$

Thus,

$$y = \frac {3 \times (5 4 2 \times 1 0 ^ {- 9} m) \times (0 . 6 8 m)}{1 . 2 \times 1 0 ^ {- 4} m} = 0. 0 0 9 2 1 4 \mathrm {m} \approx 9. 2 \mathrm {m m}$$

Answer: $9.2 \mathrm{~mm}$

Source: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.

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