In YDSE, $d = 5\lambda $ then the total number of maxima observed on screen will be
(A) $9$
(B) $8$
(C) $7$
(D) $5$

Hint: We know that in YDSE the path difference is given by
$\Delta x = n\lambda $ Where, $\lambda $ is the wavelength of light and in terms of angular width path difference is given by $\Delta x = d\sin \theta $ where $\theta $ is angular width of fringe and $d$ is distance between two slits. We will equate both the equations and solve for the required result.

Complete step by step solution:
Given: $d = 5\lambda $
Now we know that the angular width $\theta = \dfrac{\lambda }{d}......(1)$
Since the maximum angle can be ${90^ \circ }$ therefore,
Number of fringes, $n = \dfrac{{\sin 90}}{{\sin \theta }}......(2)$
But as $\theta $ is small, so we can write $\sin \theta \approx \theta $
Hence equation (2) can be written as
$n = \dfrac{1}{\theta }......(3)$
Now from equation (1) and (3), we can write
$n = \dfrac{d}{\lambda }$
We have given $d = 5\lambda $ hence above equation gives
$n = 5$
Hence the number of maxima is equal to
 $
   = 2n - 1 \\
   = (2 \times 5) - 1 \\
   = 9 \\
 $

Hence option (A) is correct.

Note: We know as the separation between the slits is increased, the fringe width is decreased. If $d$ becomes much smaller than $\lambda $ the fringe width will be very small. The maxima and minima in this case will be so closely spaced that it will look like a uniform intensity pattern. This is an example of a general result that the wave effects are difficult to observe, if the wavelength is small compared to the dimensions of the obstructions or openings to the incident wavefront.