Question

In YDSE,the slits are 2 mm apart and are illuminated by photons of 2 wave length 12000 Ao and 10000 Ao. At what min. Distance from the common central bright fringe on the screen 2m from the slit will a bright from one interference pattern coincide with bright fringe from other ?

Answer 1

Answer on Question#40905, Physics, Optics

In YDSE, the slits are $2\mathrm{mm}$ apart and are illuminated by photons of 2 wave length $12000\mathring{\mathrm{A}}$ and $10000\mathring{\mathrm{A}}$ . At what minimal distance from the common central bright fringe on the screen $2\mathrm{m}$ from the slit will a bright from one interference pattern coincide with bright fringe from other?

Solution:

The condition for maximum (bright spot) is

$$d \sin \theta = m \lambda$$

where $m$ is order of interference, $D = 2m$ , $d = 2.0 \times 10^{-3}m$ .

Here as we are considering the coincidence of two bright fringes, that is why the value of $\sin \theta$ will be the same for both.

Let $m_1$ is order of bright fringe for 12000 Å and $m_2$ is order of bright fringe for 10000 Å, and they will coincide.

$$d \sin \theta = m _ {1} \lambda_ {1}$$ $$d \sin \theta = m _ {2} \lambda_ {2}$$ $$m _ {1} \lambda_ {1} = m _ {2} \lambda_ {2}$$ $$\frac {m _ {2}}{m _ {1}} = \frac {\lambda_ {1}}{\lambda_ {2}} = \frac {1 2 0 0 0}{1 0 0 0 0} = \frac {6}{5}$$

So we can say that for minimum distance bright fringe number 6 for $10000\mathring{\mathrm{A}}$ will coincide with bright fringe number 5 or $12000\mathring{\mathrm{A}}$

The distance between two adjacent bright spots on the screen is

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

where $m$ is order of interference, $D = 2m$ , $d = 2.0 \times 10^{-3}m$ .

Thus,

$$y = \frac {6 \cdot 1 0 0 0 0 \cdot 1 0 ^ {- 1 0} \cdot 2}{2 \cdot 1 0 ^ {- 3}} = 0. 0 0 6 \mathrm {m} = 6 \mathrm {m m}$$

Answer. $6 \mathrm{~mm}$ .