Question

Light from a mercury lamp falls on two slits separated by 0.6 mm, and the resulting interference pattern is observed on a screen 2.5 m away from the slits. If the adjacent bright fringes are separated by 2.27 mm, what is the wavelength of the light?

Answer 1

Answer on Question #67025, Physics / Optics

Light from a mercury lamp falls on two slits separated by $0.6 \, \text{mm}$ , and the resulting interference pattern is observed on a screen $2.5 \, \text{m}$ away from the slits. If the adjacent bright fringes are separated by $2.27 \, \text{mm}$ , what is the wavelength of the light?

Find: $\lambda - ?$

Given:

$d = 0.6 \times 10^{-3} \mathrm{~m}$

$L = 2.5 \mathrm{~m}$

$\Delta x = 2.27 \times 10^{-3} \mathrm{~m}$

$n = 1.0$

Solution:

$S_{1}$ and $S_{2}$ are the point sources

d is the distance between $S_{1}$ and $S_{2}$

Zero interference maximum is in point $O_1$ .

L is the distance between the point sources and screen

Choose on screen the point $M$ , where there is a dark line.

Dark line responsible the interference minimum.

Condition of interference maximum:

$\mathrm{n}\Delta \mathrm{r} = \mathrm{k}\lambda (1),$

where $\Delta r$ is the geometric difference at which the waves come to the point $M$ from $S_{1}$ and $S_{2}$ , $n$ is the absolute index of refraction, $\lambda$ is the wavelength of light, $k = 0, \pm 1, \pm 2, \ldots$ (the number of minimum)

From Figure $\Rightarrow$ the similarity of triangles (by two angles):

$\Delta O O_{1} M \sim \Delta S_{1} N S_{2}$

$$\frac {S _ {1} S _ {2}}{M O} = \frac {S _ {2} N}{O _ {1} M}$$

Of (2) and Figure $\Rightarrow \frac{\mathrm{d}}{\mathrm{MO}} = \frac{\Delta r}{x}$ (3)

Since $d << L$ , then $MO \approx L$ (4)

(4) in (3): $\frac{d}{L} = \frac{\Delta r}{x}$ (5)

Of (5) $\Rightarrow \Delta r = \frac{d}{L} x$ (6)

(6) in (1): $n \frac{d}{L} x_k = k\lambda$ (7)

Of (7) $\Rightarrow x_{k} = \frac{k\lambda L}{nd}$ (8)

Of (7) $\Rightarrow x_{k + 1} = \frac{(k + 1)\lambda L}{nd}$ (9)

$\Delta x = x_{k + 1} - x_k$ (10)

(8) and (9) in (10): $\Delta x = \frac{\lambda L}{nd}$ (11)

Of (11) $\Rightarrow \lambda = \frac{\mathrm{nd}\Delta\mathrm{x}}{\mathrm{L}}$ (12)

Of (12) $\Rightarrow \lambda = 544.8 \times 10^{-9} \mathrm{~m}$

**Answer:**

$544.8 \times 10^{-9} \mathrm{~m}$

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