In a Young’s Double slit experiment, the distance between the two identical slits is 6.1 times larger than the slit width. The number of intensity maxima observed within the central maximum of the single slit diffraction pattern is:
A. 3
B. 6
C. 12
D. 24

Hint:Young’s Double slit experiment demonstrates that light has wave nature.
Width of maxima in Single Slit diffraction experiment is given by the formula:
${w_s} = 2(\dfrac{{D\lambda }}{a})$
Where,
\[{w_s}\] is the width of maxima in single slit diffraction experiment
D is the distance between slit and screen
a is the aperture of the slit
$\lambda $ is the wavelength of light
Width of the central maxima in Young’s Double Slit experiment is given by the formula:
${w_d} = \dfrac{{D\lambda }}{d}$
Where,
\[{w_d}\] is the width of maxima in Young’s double slit experiment
D is the distance between slit and screen
d is the distance between two slits
$\lambda $ is the wavelength of light


Complete step by step solution:
Width of maxima in Single Slit diffraction experiment is given by the formula:
${w_s} = 2(\dfrac{{D\lambda }}{a})$
Where,
\[{w_s}\] is the width of maxima in single slit diffraction experiment
D is the distance between slit and screen
a is the aperture of the slit
$\lambda $ is the wavelength of light
Width of the central maxima in Young’s Double Slit experiment is given by the formula:
${w_d} = \dfrac{{D\lambda }}{d}$
Where,
\[{w_d}\] is the width of maxima in Young’s double slit experiment
D is the distance between slit and screen
d is the distance between two slits
$\lambda $ is the wavelength of light
Now, let us assume that there are n maxima lying within the central maxima. Then n times the width of central maxima of Young’s Double Slit experiment must be equal to width of maxima in Single Slit experiment.
This can be established by:
$ = > n(\dfrac{{D\lambda }}{d}) = 2(\dfrac{{D\lambda }}{a})$
Cancelling $D\lambda $ from both sides in the above equation,
$ = > n(\dfrac{1}{d}) = \dfrac{2}{a}$
Now, it is given in question that the distance between the two identical slits is 6.1 times larger than the slit width. This clearly implies that,
$d = 6.1 \times a$
Inserting the value of d in the above equation,
We get,
$ = > n(\dfrac{1}{{6.1 \times a}}) = \dfrac{2}{a}$
Cancelling a on both sides,
$ = > \dfrac{n}{{6.1}} = 2$
$ = > n = 12.2$
$ = > n \approx 12$

Hence Option (C) is correct.

Note:One must have a clear conceptual understanding in order to solve such tricky questions. Additionally, one must be able to recall formulas of Young’s Double Slit Experiment and Single Slit Diffraction Experiment.