Question

in YDSE how many maxima can be obtained on the screen if the wavelength of light used is $200\,nm$ and $d = 700\,nm$
A. $12$
B. $7$
C. $18$
D. None of these

Answer 1

Here, YDSE is abbreviated as young’s double-slit experiment. Young’s double-slit experiment is defined as proof of the wave theory of light. Here, we will use the formula of fringe width to calculate the angular width of the fringes of light. Also, we will calculate the number of fringes formed to calculate the maxima of fringes obtained on the screen.

Formula used:
The formula used for fringe width is given by

$\theta = \dfrac{\lambda }{d}$

Here $\theta$ is the angular width of the fringe, $\lambda$ is the wavelength of the light and $d$ is the distance between the two slits.

Also, the formula of the number of fringes is given by
$n = \dfrac{{\sin {\theta _2}}}{{\sin {\theta _1}}}$

Complete step by step answer:
In the given question, the wavelength of light is $\lambda = 200\,nm$
And the distance between the two slits is $d = 700\,nm$
The fringe width is given by

$\theta = \dfrac{{200}}{{700}}$
$\Rightarrow \,\theta = \dfrac{2}{7}$

Now, as we know that the maximum angle made by the light is $90^\circ$ .

Now, if $n$ is the number of fringes. Therefore, the number of fringes is given by

$n = \dfrac{{\sin 90^\circ }}{{\sin \theta }}$
$\Rightarrow n = \dfrac{1}{{\sin \theta }}$

Now, as $\sin \theta$ is very small, therefore, we can replace $\sin \theta = \theta$.
Therefore, the above relation will become
$n = \dfrac{1}{\theta }$
Putting the value of $\theta$ in the above equation
$n = \dfrac{7}{2}$
Now, the number of maxima and minima of the slits can be calculated by using the following formula $N = 2n$

Now, putting the value of $n$ in the above equation, we get
$N = 2 \times \dfrac{7}{2}$
$\Rightarrow \,N = 7$
Therefore, in YDSE, there is a $7$ maxima that can be obtained on the screen.

Hence, option (B) is the correct option.

Note: In the double-slit experiment, when the light is through two adjacent slits, an interference pattern will be observed. This interference pattern is the result of the wave nature of the light. Here, in the above example, we have taken $\theta = 90^\circ$ because the maximum angle made by the light is $90^\circ