Wave Optics — NCERT Solutions

In YDSE, position of the $n$th bright fringe is
$$x_n = \frac{n\lambda D}{d}$$
Given:

• $d = 0.28\,\text{mm} = 0.28\times10^{-3}\,\text{m}$
• $D = 1.4\,\text{m}$
• $x_4 = 1.2\,\text{cm} = 1.2\times10^{-2}\,\text{m}$
• $n=4$

So,
$$\lambda = \frac{x_nd}{nD}$$
$$\lambda = \frac{(1.2\times10^{-2})(0.28\times10^{-3})}{4\times1.4}$$
$$= \frac{3.36\times10^{-6}}{5.6} = 6.0\times10^{-7}\,\text{m}$$
Thus the wavelength is $6.0\times10^{-7}\,\text{m}$ or 600 nm.

For two coherent sources, the intensity is
$$I = 4I_0\cos^2\left(\frac{\phi}{2}\right)$$
A path difference of $\lambda$ means phase difference
$$\phi = 2\pi$$
so the waves are in phase and the intensity is maximum:
$$I = 4I_0$$
Given that at path difference $\lambda$, the intensity is $K$, we have
$$K = 4I_0$$
For path difference $\lambda/3$, phase difference is
$$\phi = 2\pi\left(\frac{1}{3}\right)=\frac{2\pi}{3}$$
Hence,
$$I = 4I_0\cos^2\left(\frac{\pi}{3}\right)=4I_0\left(\frac{1}{2}\right)^2 = I_0$$
Since $K = 4I_0$,
$$I = \frac{K}{4}$$
So the intensity is $K/4$.

For Young’s double-slit experiment, the position of the $n$th bright fringe is

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

From the chapter, the two wavelengths are $\lambda_1 = 650\,\text{nm}$ and $\lambda_2 = 520\,\text{nm}$.

(a) Third bright fringe for $650\,\text{nm}$

For the third bright fringe, $n=3$:

$$x_3 = \frac{3\lambda D}{d}$$

So its distance from the central maximum is

$$x_3 = \frac{3 \times 650\,\text{nm} \times D}{d}$$

or, in terms of the setup,

$$x_3 = \frac{1950\,\text{nm} \cdot D}{d} = \frac{1.95\times 10^{-6}\,\text{m}\cdot D}{d}$$

(b) Least distance where bright fringes coincide

Bright fringes of both wavelengths coincide when their path differences are equal to whole-number multiples of both wavelengths. So we need the least common multiple of $650$ nm and $520$ nm.

Prime factors:

$$650 = 2 \times 5^2 \times 13$$
$$520 = 2^3 \times 5 \times 13$$

So

$$\text{LCM} = 2^3 \times 5^2 \times 13 = 2600\,\text{nm}$$

Thus the least common bright fringe occurs when the path difference is $2600$ nm. The corresponding distance from the central maximum is

$$x = \frac{m\lambda D}{d}$$

using either wavelength:

For $650$ nm, $m=4$;
for $520$ nm, $m=5$.

Hence the least distance from the central maximum where the bright fringes coincide is the position of the 4th bright fringe for 650 nm or 5th bright fringe for 520 nm:

$$x = \frac{4\times 650\,\text{nm}\times D}{d} = \frac{5\times 520\,\text{nm}\times D}{d} = \frac{2600\,\text{nm}\cdot D}{d}$$

So the coincidence occurs at the distance corresponding to path difference 2600 nm.