Question

The different arrangements of incident beam and position of glass slab in a $\text{YDSE}$ experiment have been shown in the figure.

$\left(a\right)$	$\left(p\right)$ The zero order maxima will lie above point $O$.
$\left(b\right)$	$\left(q\right)$ The zero order maxima will lie below point $O.$
$\left(c\right)$	$\left(r\right)$ The zero order maxim a may lie above or below point $O$.
$\left(d\right)$	$\left(s\right)$ The zero order maxima may lie at point $O$.

Select the correct option.

• A. $a\to q,b\to p,c\to r,d\to s$
• B. $a\to p,b\to r,s,c\to p,d\to p$
• C. $a\to p,b\to p,c\to r,s,d\to s$
• D. $a\to q,b\to r,c\to s,d\to p$

Answer 1

The correct option is B $a\to p,b\to r,s,c\to p,d\to p$

$\text{Case (a)}:$ The incident beam on the slits in inclined. The inclined beam of light and introduction of glass slab in front of the upper-slit will shift the fringe pattern upwards altogether.

So, zero order maxima will lie above $O$.


$\text{Case (b):}$ The inclined beam of light will tend to shift the fringe pattern upwards. However, presence of slab near the lower slit will tend to shift the fringe pattern downwards. Hence, zero order maxima may lie above $O$ or below $O$, or at point $O$.

$\text{Case (c)}:$ The inclined beam of light will produce an upward shift in fringe pattern.

$\text{Case (d)}:$ the central maxima (zero order) will form at the point where path difference between interfering waves is zero. i.e $\Delta x=0$

Point $O$ is not at central position, hence $\Delta x\ne 0$ at $O$. Thus, fringe pattern gets shifted upward.

Hence, $\left(B\right)$ is the correct answer.

Why this question ? Note: Position of central maxima is determined by net path difference $=0$, caused by combination of inclined beam/ beam orientation and slab position. If the central maxima shift upward, the whole fringe pattern gets an upward shift.