Question

In YDSE amplitude of wave depends directly on width of the slit. Intensity of light from lower slit is $l_0$

List-I	List-II
(P) If upper slit is of double width of lower slit then ratio of minimum to maximum intensity is	(1) $\frac{1}{3}$
(Q) If there is third slit of twice width between two identical slits then ratio of minimum to maximum intensity is	(2) $\frac{1}{4}$
(R) If upper slit is 1.5 times wider than lower slit then average intensity of pattern is $\frac{I_0}{K}$, then K is	(3) $\frac{4}{13}$
(S) If there is third slit of same width between two identical slits (each intensity $l_0$) then maximum intensity is $\frac{I_0}{l}$ then /is	(4) Zero
	(5) $\frac{1}{9}$

• A. P-5, Q-4, R-3, S-5
• B. P-1, Q-2, R-3, S-4
• C. P-5, Q-1, R-2, S-3
• D. P-4, Q-5, R-3, S-1

Answer 1

Answer: (A)

P-5, Q-4, R-3, S-5

Answer 2

The problem describes Young's Double Slit Experiment (YDSE) where the amplitude of the wave is directly proportional to the width of the slit. This implies that the intensity ($I$) is proportional to the square of the amplitude ($A^2$), and thus proportional to the square of the slit width ($w^2$). We can write $I \propto w^2$. Let $I_1$ and $I_2$ be the intensities from two slits with amplitudes $A_1$ and $A_2$. Then $I_1 = A_1^2$ and $I_2 = A_2^2$. If $w_1$ and $w_2$ are the widths, then $A_1 = cw_1$ and $A_2 = cw_2$, leading to $I_1 = c^2w_1^2$ and $I_2 = c^2w_2^2$. The ratio of intensities is $\frac{I_1}{I_2} = \frac{w_1^2}{w_2^2}$.

The intensity at any point in the interference pattern is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi$, where $\phi$ is the phase difference. Maximum intensity ($I_{max}$) occurs when $\cos\phi = 1$, so $I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} = (\sqrt{I_1} + \sqrt{I_2})^2 = (A_1 + A_2)^2$. Minimum intensity ($I_{min}$) occurs when $\cos\phi = -1$, so $I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = (\sqrt{I_1} - \sqrt{I_2})^2 = (A_1 - A_2)^2$. The ratio of minimum to maximum intensity is $\frac{I_{min}}{I_{max}} = \frac{(\sqrt{I_1} - \sqrt{I_2})^2}{(\sqrt{I_1} + \sqrt{I_2})^2}$. The average intensity $\langle I \rangle$ for a two-slit experiment is $\langle I \rangle = I_1 + I_2$, because the average value of $\cos\phi$ over a full cycle is zero.

We assume "Intensity of light from lower slit is $I_0$" means that for one of the slits (say slit 1), $I_1 = I_0$.

(P) If upper slit is of double width of lower slit: Let $w_1$ be the width of the lower slit and $w_2$ be the width of the upper slit. Given $w_2 = 2w_1$. Since $I \propto w^2$, $I_2 = (\frac{w_2}{w_1})^2 I_1 = (\frac{2w_1}{w_1})^2 I_1 = 4I_1$. Given $I_1 = I_0$, so $I_2 = 4I_0$. $\frac{I_{min}}{I_{max}} = \frac{(\sqrt{I_0} - \sqrt{4I_0})^2}{(\sqrt{I_0} + \sqrt{4I_0})^2} = \frac{(\sqrt{I_0} - 2\sqrt{I_0})^2}{(\sqrt{I_0} + 2\sqrt{I_0})^2} = \frac{(-\sqrt{I_0})^2}{(3\sqrt{I_0})^2} = \frac{I_0}{9I_0} = \frac{1}{9}$. Thus, (P) matches with (5).

(Q) If there is third slit of twice width between two identical slits: This is a three-slit experiment. Let the slits be S1, S2, S3. Let S1 and S3 be identical, so $w_1 = w_3 = w$. Let their intensity be $I_1 = I_3 = I$. The third slit S2 has twice the width, $w_2 = 2w$. Its intensity is $I_2 = (\frac{2w}{w})^2 I = 4I$. Assuming $I_1 = I_0$, then $I_3 = I_0$ and $I_2 = 4I_0$. The corresponding amplitudes are $A_1 = \sqrt{I_0}$, $A_3 = \sqrt{I_0}$, and $A_2 = \sqrt{4I_0} = 2\sqrt{I_0}$. For maximum intensity, all waves must be in phase. If the slits are equally spaced, this occurs at the central point. $A_{max} = A_1 + A_2 + A_3 = \sqrt{I_0} + 2\sqrt{I_0} + \sqrt{I_0} = 4\sqrt{I_0}$. $I_{max} = (4\sqrt{I_0})^2 = 16I_0$. For minimum intensity, we consider the resultant amplitude $A_{res} = A_1 + A_2 e^{i\Delta\phi} + A_3 e^{i2\Delta\phi}$, where $\Delta\phi$ is the phase difference between adjacent slits. $A_{res} = \sqrt{I_0} + 2\sqrt{I_0} e^{i\Delta\phi} + \sqrt{I_0} e^{i2\Delta\phi} = \sqrt{I_0}(1 + 2e^{i\Delta\phi} + e^{i2\Delta\phi}) = \sqrt{I_0}(1+e^{i\Delta\phi})^2$. The resultant amplitude is zero if $1+e^{i\Delta\phi} = 0$, which means $e^{i\Delta\phi} = -1$, or $\Delta\phi = \pi$. If $\Delta\phi = \pi$, then $A_{res} = \sqrt{I_0}(1+(-1))^2 = 0$. Thus, the minimum intensity $I_{min} = 0$. The ratio $\frac{I_{min}}{I_{max}} = \frac{0}{16I_0} = 0$. Thus, (Q) matches with (4).

(R) If upper slit is 1.5 times wider than lower slit then average intensity of pattern is $\frac{I_0}{K}$: Let $w_1$ be the width of the lower slit and $w_2$ be the width of the upper slit. Given $w_2 = 1.5 w_1 = \frac{3}{2} w_1$. Given $I_1 = I_0$. $I_2 = (\frac{w_2}{w_1})^2 I_1 = (\frac{3/2 w_1}{w_1})^2 I_0 = (\frac{3}{2})^2 I_0 = \frac{9}{4} I_0$. The average intensity $\langle I \rangle = I_1 + I_2 = I_0 + \frac{9}{4} I_0 = \frac{13}{4} I_0$. The question states $\langle I \rangle = \frac{I_0}{K}$. So, $\frac{13}{4} I_0 = \frac{I_0}{K}$. $K = \frac{4}{13}$. Thus, (R) matches with (3).

(S) If there is third slit of same width between two identical slits (each intensity $l_0$) then maximum intensity is $\frac{I_0}{l}$: This is a three-slit experiment. Let the slits be S1, S2, S3. S1 and S3 are identical with intensity $I_1 = I_3 = I_0$. S2 is between them and has the same width, so $w_2 = w_1 = w_3$. Thus $I_2 = I_1 = I_3 = I_0$. The amplitudes are $A_1 = \sqrt{I_0}$, $A_2 = \sqrt{I_0}$, $A_3 = \sqrt{I_0}$. For maximum intensity, all waves are in phase (at the central maximum). $A_{max} = A_1 + A_2 + A_3 = \sqrt{I_0} + \sqrt{I_0} + \sqrt{I_0} = 3\sqrt{I_0}$. $I_{max} = (3\sqrt{I_0})^2 = 9I_0$. The question states $I_{max} = \frac{I_0}{l}$. So, $9I_0 = \frac{I_0}{l}$. $l = \frac{1}{9}$. Thus, (S) matches with (5).

The correct matches are: (P) - (5) (Q) - (4) (R) - (3) (S) - (5)