In a Young's double slit experiment, two slits are illuminated with a light of wavelength $$800$$ nm. The line joining $$A_1P$$ is perpendicular to $$A_1A_2$$ as shown in the figure. If the first minimum is detected at $$P$$, the value of slits separation $$a$$ will be:

The distance of screen from slits $$D = 5$$ cm. Answer in mm.

Solution :

For first minima in Young’s double slit experiment :

$$\text{Path difference} = \frac{\lambda}{2}$$

At point P,

$$A_1P = D$$

$$A_2P = \sqrt{D^2+a^2}$$

Therefore,

$$\sqrt{D^2+a^2} - D = \frac{\lambda}{2}$$

Given :

$$\lambda = 800\text{ nm} = 8 \times 10^{-7}\text{ m}$$

$$D = 5\text{ cm} = 5 \times 10^{-2}\text{ m}$$

Substituting :

$$\sqrt{D^2+a^2} = D + \frac{\lambda}{2}$$

Squaring both sides :

$$D^2+a^2 = D^2 + D\lambda + \frac{\lambda^2}{4}$$

Since $$\ \ lambda^2$$ is very small, neglecting it :

$$a^2 = D\lambda$$

$$= (5 \times 10^{-2})(8 \times 10^{-7})$$

$$= 4 \times 10^{-8}$$

$$a = 2 \times 10^{-4}\text{ m}$$

$$= 0.2\text{ mm}$$

Final Answer :

$$0.2\text{ mm}$$