Question

In a Young's double slit experiment, a plane monochromatic wave of wavelength 6000 Å, is incident normally on the slit plane as shown in the figure. A perfectly transparent film of thickness t, and refractive index 1.5 is placed in front of the slit S₁. The intensity of light on the screen near O is

I due to each slit.

• A. Value of $\frac{I}{I_0}$ at A, if t = 0.6 µm
• B. Value of $\frac{I}{I_0}$ at A, if t = 1.2 µm
• C. The minimum value of t (µm). (t > 0) for intensity to be maximum at O and match $\frac{10t}{4}$.
• D. The minimum value of t (µm). (t > 0) for intensity to be the minimum at O and match $\frac{10t}{2}$.
• E. 4
• F. 1
• G. 0
• H. 3
• I. 6

Answer 1

P-3, Q-1, R-4, S-4

Answer 2

The intensity at a point on the screen in a Young's double-slit experiment is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$, where $I_0$ is the intensity due to each slit and $\phi$ is the phase difference between the waves from the two slits at that point.

The phase difference is given by $\phi = \frac{2\pi}{\lambda} \Delta x$, where $\Delta x$ is the path difference.

In the presence of a film of thickness $t$ and refractive index $\mu$ in front of slit S₁, the additional path difference introduced is $(\mu - 1)t$.

The distance between the slits is $d = 10 \text{ mm} = 10 \times 10^{-3} \text{ m}$. The distance between the slits and the screen is $D = 2 \text{ m}$. The wavelength of light is $\lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m} = 0.6 \text{ µm}$. The refractive index of the film is $\mu = 1.5$. The additional path difference due to the film is $(\mu - 1)t = (1.5 - 1)t = 0.5t$.

Let A be a point on the screen at a distance $y_A = 0.48 \text{ mm} = 0.48 \times 10^{-3} \text{ m}$ from the central point O. The path difference at A without the film is $\Delta x_{air} = \frac{y_A d}{D} = \frac{(0.48 \times 10^{-3})(10 \times 10^{-3})}{2} = 2.4 \times 10^{-6} \text{ m}$. In terms of wavelength, $\Delta x_{air} = \frac{2.4 \times 10^{-6}}{6 \times 10^{-7}} \lambda = 4\lambda$. The total path difference at A with the film in front of S₁ is $\Delta x_A = \Delta x_{air} - (\mu - 1)t = 4\lambda - 0.5t$. The phase difference at A is $\phi_A = \frac{2\pi}{\lambda} \Delta x_A = \frac{2\pi}{\lambda} (4\lambda - 0.5t) = 8\pi - \frac{\pi t}{\lambda}$. The intensity at A is $I_A = 4I_0 \cos^2(\frac{\phi_A}{2}) = 4I_0 \cos^2(\frac{8\pi - \frac{\pi t}{\lambda}}{2}) = 4I_0 \cos^2(4\pi - \frac{\pi t}{2\lambda}) = 4I_0 \cos^2(\frac{\pi t}{2\lambda})$.

(P) Value of $\frac{I_A}{I_0}$ at A, if $t = 0.6 \text{ µm}$. $\frac{I_A}{I_0} = 4 \cos^2(\frac{\pi t}{2\lambda})$. Given $t = 0.6 \text{ µm}$ and $\lambda = 0.6 \text{ µm}$. $\frac{\pi t}{2\lambda} = \frac{\pi (0.6 \text{ µm})}{2 (0.6 \text{ µm})} = \frac{\pi}{2}$. $\frac{I_A}{I_0} = 4 \cos^2(\frac{\pi}{2}) = 4 \times 0^2 = 0$. (P) matches with (3).

(Q) Value of $\frac{I_A}{I_0}$ at A, if $t = 1.2 \text{ µm}$. Given $t = 1.2 \text{ µm}$ and $\lambda = 0.6 \text{ µm}$. $\frac{\pi t}{2\lambda} = \frac{\pi (1.2 \text{ µm})}{2 (0.6 \text{ µm})} = \frac{1.2\pi}{1.2} = \pi$. $\frac{I_A}{I_0} = 4 \cos^2(\pi) = 4 \times (-1)^2 = 4$. (Q) matches with (1).

(R) The minimum value of $t$ (µm), ($t > 0$) for intensity to be maximum at O. At point O, $y_O = 0$. The path difference at O without the film is 0. The total path difference at O with the film is $\Delta x_O = 0 - (\mu - 1)t = -0.5t$. The phase difference at O is $\phi_O = \frac{2\pi}{\lambda} (-0.5t) = -\frac{\pi t}{\lambda}$. The intensity at O is $I_O = 4I_0 \cos^2(\frac{\phi_O}{2}) = 4I_0 \cos^2(-\frac{\pi t}{2\lambda}) = 4I_0 \cos^2(\frac{\pi t}{2\lambda})$. For maximum intensity at O, $\cos^2(\frac{\pi t}{2\lambda}) = 1$. This occurs when $\frac{\pi t}{2\lambda} = n\pi$, where $n$ is an integer. $t = 2n\lambda$. For minimum $t > 0$, we take $n=1$. $t_{min} = 2\lambda = 2 \times 0.6 \text{ µm} = 1.2 \text{ µm}$. We need to match $\frac{10t}{4}$. $\frac{10 t_{min}}{4} = \frac{10 \times 1.2}{4} = \frac{12}{4} = 3$. (R) matches with (4).

(S) The minimum value of $t$ (µm), ($t > 0$) for intensity to be minimum at O. For minimum intensity at O, $\cos^2(\frac{\pi t}{2\lambda}) = 0$. This occurs when $\frac{\pi t}{2\lambda} = (n + \frac{1}{2})\pi$, where $n$ is an integer. $t = (2n + 1)\lambda$. For minimum $t > 0$, we take $n=0$. $t_{min} = (2(0) + 1)\lambda = \lambda = 0.6 \text{ µm}$. We need to match $\frac{10t}{2}$. $\frac{10 t_{min}}{2} = \frac{10 \times 0.6}{2} = \frac{6}{2} = 3$. (S) matches with (4).

Matches: (P) -> (3) (Q) -> (1) (R) -> (4) (S) -> (4)

Final Answer should be a mapping from List-I to List-II. P-3, Q-1, R-4, S-4.