In a single slit experiment, a parallel beam of green light of wavelength $$550$$ nm passes through a slit of width $$0.20$$ mm. The transmitted light is collected on a screen $$100$$ cm away. The distance of first order minima from the central maximum will be $$x \times 10^{-5}$$ m. The value of $$x$$ is :

We need to find the position of the first order minima in a single slit diffraction experiment.

Recall the condition for minima in single slit diffraction

For a single slit of width $$d$$, the condition for the $$n$$-th order minima is:

$$d \sin\theta = n\lambda$$

where $$\lambda$$ is the wavelength and $$n = 1, 2, 3, \ldots$$

Use the small angle approximation

For small angles (which is valid when the slit width is much larger than the wavelength), $$\sin\theta \approx \tan\theta = \frac{y}{D}$$, where $$y$$ is the distance of the minima from the central maximum and $$D$$ is the distance to the screen.

Substituting into the minima condition:

$$d \cdot \frac{y}{D} = n\lambda$$

$$y = \frac{n\lambda D}{d}$$

Substitute the given values for the first order minimum ($$n = 1$$)

Given: $$\lambda = 550$$ nm $$= 550 \times 10^{-9}$$ m, $$D = 100$$ cm $$= 1$$ m, $$d = 0.20$$ mm $$= 0.20 \times 10^{-3}$$ m $$= 2 \times 10^{-4}$$ m.

$$y = \frac{1 \times 550 \times 10^{-9} \times 1}{2 \times 10^{-4}}$$

$$y = \frac{550 \times 10^{-9}}{2 \times 10^{-4}} = 275 \times 10^{-5} \text{ m}$$

Therefore, $$x = 275$$.

The answer is 275.