Question

In a Young’s double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength $\lambda = 500nm$ is incident on the slits. The total number of bright fringes that are observed in the angular range $ - {30^ \circ } \leqslant \theta \leqslant {30^ \circ }$ is:
A. 320
B. 641
C. 321
D. 640

Answer 1

Hint – In the Young’s double slit experiment, the path difference is given by the formula, $d\sin \theta = n\lambda $ , where d is the distance between the slits, n is the number of maxima and theta is the angle.
Formula used - $d\sin \theta = n\lambda $

Complete step-by-step answer:
We have been given that in Young’s double slit experiment, the slits are placed 0.320 mm apart.
Also, the wavelength of light is 500 nm, and angular range is $ - {30^ \circ } \leqslant \theta \leqslant {30^ \circ }$
So, we know that the path difference is given by the formula-
$d\sin \theta = n\lambda $
So, according to question, $d = 0.320mm,\sin \theta = \sin {30^ \circ },\lambda = 500nm$
Putting these values, we can find the value of n, which is the number of maxima.
\[
  d\sin \theta = n\lambda \\
   \Rightarrow n = \dfrac{{d\sin \theta }}{\lambda } \\
   \Rightarrow n = \dfrac{{0.32 \times {{10}^{ - 3}}\sin {{30}^ \circ }}}{{500 \times {{10}^{ - 9}}}} = \dfrac{{0.32 \times {{10}^{ - 3}} \times \dfrac{1}{2}}}{{500 \times {{10}^{ - 9}}}} = \dfrac{{0.32 \times {{10}^6}}}{{500 \times 2}} = 320 \\
 \]
Now this value of n represents the no. of maxima which are for angle 30 degrees and formed above.
So, the same number of maxima will form below also and one fringe will form in the centre.

Refer the figure below for better understanding-

So, total number of maxima $ = n + 1 + n = 320 + 1 + 320 = 641$
Hence, the correct option is B.

Note – Whenever such types of questions appear, then always write the things given in the question and then as mentioned in the solution, use the formula for path difference to find the value of maxima. Also, draw a figure to understand in a better way. So, the total number of maxima is the number of maxima above + number of maxima at the centre + number of maxima below.