Question

Two slits are spaced 0.3 mm apart and are placed 50cm from a screen. Suppose the entire apparatus is immersed in water. What is the distance between the second and third dark lines of the interference pattern when the slits are illuminated with the light of 600 nm wavelength?

Answer 1

Answer on Question #66908, Physics / Optics

Two slits are spaced $0.3\mathrm{mm}$ apart and are placed $50\mathrm{cm}$ from a screen. Suppose the entire apparatus is immersed in water. What is the distance between the second and third dark lines of the interference pattern when the slits are illuminated with the light of $600\mathrm{nm}$ wavelength?

Find: $\Delta x - ?$

Given:

$d = 0.3 \times 10^{-3} \mathrm{~m}$

$L = 0.5 \mathrm{~m}$

$n = 1.33$

$\lambda = 600 \times 10^{-9} \mathrm{~m}$

$S_{1}$ 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 minimum:

$\mathrm{n}\Delta \mathrm{r} = (2\mathrm{k} + 1)\frac{\lambda}{2} (1),$

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):

$\Delta \mathrm{OO}_1\mathrm{M}\sim \Delta \mathrm{S}_1\mathrm{NS}_2$

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

Of (2) and Figure $\Rightarrow \frac{d}{MO} = \frac{\Delta r}{x}$ (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): $n \frac{d}{L} x_k = (2k + 1) \frac{\lambda}{2}$ (7)

Of (7) $\Rightarrow x_{k} = \frac{(2k + 1)\frac{\lambda}{2}}{n_{L}^{d}}$ (8)

Of (8) $\Rightarrow x_{k} = \frac{(2k + 1)\lambda L}{2nd}$ (9)

Of (9) $\Rightarrow x_{k + 1} = \frac{(2(k + 1) + 1)\lambda L}{2nd}$ (10)

$\Delta x = x_{k + 1} - x_{k}$ (11)

(9) and (10) in (11): $\Delta x = \frac{\lambda L}{nd}$ (12)

Of (12) $\Rightarrow \Delta x = 0.752 \times 10^{-3} \, \text{m}$

**Answer:**

$0.752 \times 10^{-3} \, \text{m}$

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