Question

In a Young's double slit experiment, the separation between the slits = 2⋅0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2⋅0 m. If the intensity at the centre of the central maximum is 0⋅20 W m⁻², what will be the intensity at a point 0⋅5 cm away from this centre along the width of the fringes?

Answer 1

Given:
Separation between the slits, $d=2\mathrm{mm}=2\times {10}^{-3}\mathrm{m}$
Wavelength of the light, $\lambda =600\mathrm{nm}=6\times {10}^{-7}\mathrm{m}$
Distance of the screen from the slits, D = 2⋅0 m
$$\begin{gathered} I_{\max }=0.20\mathrm{W}/{\mathrm{m}}^2 \\ \mathrm{For}\mathrm{the}\mathrm{point}\mathrm{at}\mathrm{a}\mathrm{position}y=0.5\mathrm{cm}=0.5\times {10}^{-2}\mathrm{m}, \\ \text{p}\mathrm{ath}\mathrm{difference},∆x=\frac{yd}{D}. \\ \Rightarrow ∆x=\frac{0.5\times {10}^{-2}\times 2\times {10}^{-3}}{2} \\ =5\times {10}^{-6}\mathrm{m} \end{gathered}$$

So, the corresponding phase difference is given by
$$\begin{gathered} ∆\varphi =\frac{2\pi ∆x}{\lambda }=\frac{2\pi \times 5\times {10}^{-6}}{6\times {10}^{-7}} \\ =\frac{50\pi }{3}=16\pi +\frac{2\pi }{3} \\ \mathrm{or}∆\mathrm{\varphi }=\frac{2\pi }{3} \end{gathered}$$
So, the amplitude of the resulting wave at point y = 0.5 cm is given by
$$\begin{gathered} A=\sqrt{a^2+a^2+2a^2\cos \left(\frac{2\pi }{3}\right)} \\ =\sqrt{a^2+a^2-a^2}=a \end{gathered}$$
Similarly, the amplitude of the resulting wave at the centre is 2a.
Let the intensity of the resulting wave at point y = 0.5 cm be I.
$$\begin{gathered} \mathrm{Since}\frac{I}{I_{\max }}=\frac{A^2}{{\left(2a\right)}^2},\mathrm{we}\mathrm{have}: \\ \frac{I}{0.2}=\frac{A^2}{4a^2}=\frac{a^2}{4a^2} \\ \Rightarrow I=\frac{0.2}{4}=0.05\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$
Thus, the intensity at a point 0.5 cm away from the centre along the width of the fringes is 0.05 W/m².