Question 263: Wave Optics

Question

In Young's double slit experiment, when light of wavelength 600 nm is used, 18 fringes are observed on the screen. If the wavelength of light is changed to 400 nm , the number of fringes observed on the screen is

Accepted Answer

Let the distance between the slits and the screen be $D$ and slit separation be $d$. The width of the screen that is illuminated and on which fringes are observed is $L$.
The fringe width
(distance between two consecutive bright or dark fringes) is given by:
$ \beta = \frac{\lambda D}{d} $
The
total number of fringes ($n$)
that can be seen on the screen (of length $L$) is:
$ n = \frac{L}{\beta} $
So,
$ n = \frac{L}{\frac{\lambda D}{d}} = \frac{L d}{\lambda D} $
This shows that $n \propto \frac{1}{\lambda}$ if $L$, $d$, and $D$ are unchanged.
Let $n_1$ be the initial number of fringes for wavelength $\lambda_1$ and $n_2$ the number for wavelength $\lambda_2$:
$ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} $
Given:
$n_1 = 18$ for $\lambda_1 = 600 \, \mathrm{nm}$
$\lambda_2 = 400 \, \mathrm{nm}$
$n_2 = ?$
Substitute the values:
$ \frac{18}{n_2} = \frac{400}{600} $
$ \frac{18}{n_2} = \frac{2}{3} $
$ n_2 = \frac{18 	imes 3}{2} = 27 $
Correct answer:
$ \boxed{27} $
So, the correct option is
Option B: 27
.

Answer

12

Answer

27

Answer

22

Answer

24

Wave Optics

In Young's double slit experiment, when light of wavelength 600 nm is used, 18 fringes are observed on the screen. If the wavelength of light is changed to 400 nm , the number of fringes observed on the screen is

Choose the correct answer: