Question

White coherent light (4000-7000 Å) is sent through the slits of a YDSE. The separation between the slits is 0.5 mm and screen is 50 cm away from the slits. There is a hole in the screen at a point 1.0 mm away (along the width of the fringe) from the central line. (a) Which wavelength(s) will be absent in the light coming from the hole? (b) Which wavelength(s) will have a strong intensity?

Answer 1

(a) 4000 Å and 6667 Å (b) 5000 Å

Answer 2

For a YDSE, the position of bright fringes is given by $y = \frac{n\lambda D}{d}$ and dark fringes by $y = \frac{(k + 1/2)\lambda D}{d}$. Here, $d = 0.5 \text{ mm}$, $D = 0.5 \text{ m}$, and $y = 1.0 \text{ mm}$. Thus, $\frac{yd}{D} = \frac{(1.0 \times 10^{-3} \text{ m})(0.5 \times 10^{-3} \text{ m})}{0.5 \text{ m}} = 1.0 \times 10^{-6} \text{ m}^2$.

(a) For absent wavelengths (dark fringes), $\lambda = \frac{yd}{(k+1/2)D} = \frac{1.0 \times 10^{-6} \text{ m}^2}{k+1/2}$. For $4000 \text{ Å} \le \lambda \le 7000 \text{ Å}$, we have $4.0 \times 10^{-7} \text{ m} \le \frac{1.0 \times 10^{-6}}{k+1/2} \le 7.0 \times 10^{-7} \text{ m}$. This implies $1.428 \le k+1/2 \le 2.5$. The possible values for $k+1/2$ are $1.5$ (for $k=1$) and $2.5$ (for $k=2$). For $k+1/2 = 1.5$, $\lambda = \frac{1.0 \times 10^{-6}}{1.5} \approx 6667 \text{ Å}$. For $k+1/2 = 2.5$, $\lambda = \frac{1.0 \times 10^{-6}}{2.5} = 4000 \text{ Å}$.

(b) For strong intensity (bright fringes), $\lambda = \frac{yd}{nD} = \frac{1.0 \times 10^{-6} \text{ m}^2}{n}$. For $4000 \text{ Å} \le \lambda \le 7000 \text{ Å}$, we have $4.0 \times 10^{-7} \text{ m} \le \frac{1.0 \times 10^{-6}}{n} \le 7.0 \times 10^{-7} \text{ m}$. This implies $1.428 \le n \le 2.5$. The only integer value for $n$ is $2$. For $n=2$, $\lambda = \frac{1.0 \times 10^{-6}}{2} = 5000 \text{ Å}$.