Question

A double slit interference pattern is produced on a screen, as shown in the figure, using monochromatic light of wavelength $500nm$. Point $P$ is the location of the central bright fringe, that is produced when light waves arrive in phase without any path difference. A choice of three strips A, B and C of transparent materials with different thickness and refractive indices is available, as shown in the table. These are placed over one or both of the slits, singularly or in conjunction, causing the interference pattern to be shifted across the screen from the original pattern. In the column - I, how the strips have been placed , is mentioned whereas in the column -II, order of the fringe at point P on the screen that will be produced due to the placement of the strip (s), is shown. Correctly match both the columns.

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

Answer 1

The correct option is D $P-3,Q-3,R-4,S-1$

The shift in the central maxima due to the introduction of the glass plate is $\Delta y=\frac{t(\mu -1)D}{d}=t(\mu -1)\frac{\beta }{\lambda },$ towards the slit covered by the plate.
where, $\beta =\frac{\lambda D}{d}$ is the fringe width,
$t$ is thickness of the plate,
$\mu$ is refractive index of the plate and
$\lambda$ is the wavelength of light.
Here $\lambda =0.5\mu m$
Shift due to strip-A alone:
$\Delta y_A=5\left(\frac{1.5-1}{0.5}\right)\beta =5\beta \text{Shift due to strip-B alone:}\Delta y_B=1.5\left(\frac{2.5-1}{0.5}\right)\beta =4.5\beta \text{Shift due to strip-C alone:}\Delta y_C=0.25\left(\frac{2-1}{0.5}\right)\beta =0.5\beta$

$P)\text{Only strip B is placed over slit-I}\Delta y_B=4.5\beta \Rightarrow \text{As central maxima is shifted by}4.5\beta \text{, there will be fifth dark at P}Q)\text{strip A is placed over slit-I and strip C is placed over slit-II}\Delta y_{net}=\Delta y_A-\Delta y_C\Delta y_{net}=5\beta -0.5\beta \Delta y_{net}=4.5\beta \Rightarrow \text{As central maxima is shifted by}4.5\beta \text{, there will be fifth dark at P}$

$R)\text{strip A is placed over slit-I and Strip B and strip C is placed over slit-II}\Delta y_{net}=\Delta y_A-(\Delta y_B+\Delta y_C)\Delta y_{net}=5\beta -5\beta \Delta y_{net}=0\Rightarrow \text{As there is no shift of central maxima , P will remain to be central maxima}$

$S)\text{Strip-A and strip C are placed over slit-I and strip-B over slit -II}\Delta y_{net}=(\Delta y_A+\Delta y_C)-\Delta y_B\Delta y_{net}=5.5\beta -4.5\beta \Delta y_{net}=\beta \Rightarrow \text{As central maxima is shifted by}\beta \text{, there will be first maxima at P}$