Sodium light of wavelengths $$650$$ nm and $$655$$ nm is used to study diffraction at a single slit of aperture $$0.5$$ mm. The distance between the slit and the screen is $$2.0$$ m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is ______ $$\times 10^{-5}$$ m.

We use sodium light of wavelengths $$\lambda_1 = 650$$ nm and $$\lambda_2 = 655$$ nm with a single slit of aperture $$a = 0.5$$ mm $$= 0.5 \times 10^{-3}$$ m. The screen is at distance $$D = 2.0$$ m.

Concept: In single-slit diffraction, the first secondary maximum occurs approximately midway between the first and second minima. The minima are located at $$a \sin\theta = m\lambda$$. The first secondary maximum is approximately at $$a \sin\theta = \dfrac{3\lambda}{2}$$.

The position of the first secondary maximum on the screen (using small angle approximation $$\sin\theta \approx \tan\theta = \dfrac{y}{D}$$) is given by:

$$y = \dfrac{3\lambda D}{2a}$$

For $$\lambda_1 = 650$$ nm, we get

$$y_1 = \dfrac{3 \times 650 \times 10^{-9} \times 2.0}{2 \times 0.5 \times 10^{-3}} = \dfrac{3900 \times 10^{-9}}{10^{-3}} = 3900 \times 10^{-6} \text{ m}$$

For $$\lambda_2 = 655$$ nm, we get

$$y_2 = \dfrac{3 \times 655 \times 10^{-9} \times 2.0}{2 \times 0.5 \times 10^{-3}} = \dfrac{3930 \times 10^{-9}}{10^{-3}} = 3930 \times 10^{-6} \text{ m}$$

The separation between the positions of the first maxima is

$$\Delta y = y_2 - y_1 = (3930 - 3900) \times 10^{-6} = 30 \times 10^{-6} \text{ m} = 3 \times 10^{-5} \text{ m}$$

The answer is 3 $$\times 10^{-5}$$ m.