CBSE 2025, Set 3

Question

(b) (i) Two coherent light waves, each of intensity $I_0$ superpose each other and produce interference pattern on a screen. Obtain the expression for the resultant intensity at a point where the phase difference between the waves is $\phi$. Write its maximum and minimum possible values.
(ii) In a single slit diffraction experiment, the aperture of the slit is $3	ext{ mm}$ and the separation between the slit and the screen is $1.5	ext{ m}$. A monochromatic light of wavelength $600	ext{ nm}$ is normally incident on the slit. Calculate the distance of (I) first order minimum, and (II) second order maximum, from the centre of the screen.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Resultant Intensity Expression:**
Let the displacement of the two coherent waves at a point be:
$$y_1 = a \cos(\omega t)$$
$$y_2 = a \cos(\omega t + \phi)$$
where $a$ is the amplitude and $\phi$ is the phase difference.
According to the principle of superposition, the resultant displacement $y$ is:
$$y = y_1 + y_2 = a[\cos(\omega t) + \cos(\omega t + \phi)]$$
Using the trigonometric identity $\cos C + \cos D = 2 \cos(\frac{C-D}{2}) \cos(\frac{C+D}{2})$:
$$y = 2a \cos\left(\frac{\phi}{2}ight) \cos\left(\omega t + \frac{\phi}{2}ight)$$
This represents a harmonic wave with a resultant amplitude $A = 2a \cos\left(\frac{\phi}{2}ight)$.

Since Intensity $I$ is directly proportional to the square of the amplitude ($I = kA^2$):
$$I = k [2a \cos(\frac{\phi}{2})]^2 = 4ka^2 \cos^2\left(\frac{\phi}{2}ight)$$
If the intensity of individual waves is $I_0 = ka^2$, then:
$$\boxed{I = 4I_0 \cos^2\left(\frac{\phi}{2}ight)}$$
*(Alternatively, use $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$, with $I_1=I_2=I_0$, to get $2I_0(1+\cos\phi) = 4I_0 \cos^2(\phi/2)$).*

* **Maximum intensity ($I_{max}$):** Occurs when $\cos(\phi/2) = \pm 1$, so $I_{max} = \boxed{4I_0}$.
* **Minimum intensity ($I_{min}$):** Occurs when $\cos(\phi/2) = 0$, so $I_{min} = \boxed{0}$.

**(ii) Single Slit Diffraction Calculations:**
Given:
Slit width $a = 3	ext{ mm} = 3 	imes 10^{-3}	ext{ m}$
Distance to screen $D = 1.5	ext{ m}$
Wavelength $\lambda = 600	ext{ nm} = 600 	imes 10^{-9}	ext{ m}$

**(I) Position of first order minimum ($n = 1$):**
The condition for minima is $a \sin	heta = n\lambda$. For small angles, $y = \frac{n\lambda D}{a}$.
$$y_1 = \frac{1 	imes 600 	imes 10^{-9} 	imes 1.5}{3 	imes 10^{-3}}$$
$$y_1 = 300 	imes 10^{-6}	ext{ m} = \boxed{3 	imes 10^{-4}	ext{ m}}$$

**(II) Position of second order maximum ($n = 2$):**
The condition for secondary maxima is $y_n = (2n + 1)\frac{\lambda D}{2a}$.
For $n = 2$:
$$y_2 = \frac{5\lambda D}{2a} = \frac{5 	imes 600 	imes 10^{-9} 	imes 1.5}{2 	imes 3 	imes 10^{-3}}$$
$$y_2 = \frac{4500 	imes 10^{-9}}{6 	imes 10^{-3}} = 750 	imes 10^{-6}	ext{ m} = \boxed{7.5 	imes 10^{-4}	ext{ m}}$$

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