Column-I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits and . In each of these cases and where is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refra

Question

Column-I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits $S_{1}$ and $S_{2}$ . In each of these cases $S_{1} P_{0} = S_{2} P_{0}, S_{1} P_{1} - S_{2} P_{1} = \frac{λ}{4}$ and $S_{1} P_{2} - S_{2} P_{2} = \frac{λ}{3},$ where $λ$ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index $μ$ and thickness $t$ is pasted on slit $S_{2}$ . The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point $P$ on the screen from the two slits is denoted by $δ (P)$ and the intensity by $I (P)$ . Match each situation given in Column-I with the statement(s) in Column-II valid for that situation. [2009]

Smart Achievers · Accepted Answer

(2)

For path difference $λ / 4$ , phase difference is $\frac{π}{2} .$

For path difference $λ / 3$ , phase difference is $\frac{2 π}{3} .$

Here, $S_{1} P_{0} - S_{2} P_{0} = 0$

$∴$ $δ (P_{0}) = 0$

The path difference for $P_{1}$ and $P_{2}$ will not be zero. The intensities at $P_{0}$ is maximum.

Intensity continuously decreases from $P_{0}$ towards $P_{2}$ .

$∴$ $I (P_{0}) > I (P_{1})$

(B) At $P_{1}$ , path difference is zero. Hence $P_{1}$ is the brightest central fringe and $δ P_{1} = 0 .$

$δ P_{0} = \frac{λ}{4}, δ P_{1} = 0, δ P_{2} = \frac{λ}{12}$

(C) Here, $δ (P_{0}) = - \frac{λ}{2}; δ (P_{1}) = - \frac{λ}{4}; δ (P_{2}) = - \frac{λ}{6}$

$I (P_{0}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- π)$

$= I_{1} + I_{2} - 2 \sqrt{I_{1}} \sqrt{I_{2}} = I_{0} + I_{0} - 2 I_{0} = 0$

$I (P_{1}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- \frac{π}{2})$

$= I_{1} + I_{2} = I_{0} + I_{0} = 2 I_{0}$

$I (P_{2}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- \frac{π}{3})$

$= I_{1} + I_{2} + \sqrt{I_{1}} \sqrt{I_{2}} = I_{0} + I_{0} + I_{0} = 3 I_{0}$

$∴$ $I (P_{2}) > I (P_{1}) > I (P_{0})$

(D) Here, $δ P_{0} = \frac{3 λ}{4}; δ P_{1} = - \frac{λ}{2}; δ P_{2} = - \frac{5 λ}{12}$

$I (P_{0}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- \frac{3 π}{2})$

$= I_{1} + I_{2} = I_{0} + I_{0} = 2 I_{0}$

$I (P_{1}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- π)$

$= I_{1} + I_{2} - 2 \sqrt{I_{1}} \sqrt{I_{2}} = I_{0} + I_{0} - 2 \sqrt{I_{0}} \sqrt{I_{0}} = 0$

$I (P_{2}) = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos (- \frac{5 π}{6})$

$= I_{1} + I_{2} - \sqrt{3} \sqrt{I_{1}} \sqrt{I_{2}} = (2 - \sqrt{3}) I_{0}$

Clearly, $I (P_{1}) = 0$

$I (P_{0}) > I (P_{1})$ and $I (P_{2}) > I (P_{1})$

Solution

1 A - r, s, t; B - t; C - p, s; D - q

2 A - p, s; B - q; C - t; D - r, s, t

3 A - r, s, t; B - t; C - q; D - p, s

4 A - r, s, t; B - q; C - t; D - p, s

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