
Find the angular separation between the consecutive bright fringes in a Young’s double slit experiment with the blue-green light of wavelength $500\,nm.$ the separation between the slits is $2.0 \times {10^{ - 3}}\,m$.
Answer
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Hint:In wave optics, interference is a phenomenon where two waves superimpose with each other and later form wave of higher and lower amplitudes respectively when intensity of thus forming wave is observed on screen through Young’s double slit experiment, we get different intensities and thus the point where intensities is high and form bright fringes called constructive bright fringes, we will use the general formula of finding the fringe width to determine angular separation distance.
Formula used:
Fringe width in Young’s double slit experiment is calculated as,
$\beta = \dfrac{D}{{\lambda d}}$
where, $D,d$ are the distance between slit and screen, and distance between slits and
$\lambda $ Is the wavelength of the wave.
Angular width formula is given by,
$\theta = \dfrac{\beta }{D}$
Complete step by step answer:
According to the question, we have given that,
$\lambda = 500\,nm = 500 \times {10^{ - 9}}\,m$ and $d = 2.0 \times {10^{ - 3}}\,m$
Now using the formula, $\theta = \dfrac{\beta }{D}$ we have,
$\theta = \dfrac{{D\lambda }}{{dD}}$
$\Rightarrow \theta = \dfrac{\lambda }{d}$
On putting the values we get,
$\theta = \dfrac{{500 \times {{10}^{ - 9}}}}{{2 \times {{10}^{ - 3}}}}$
$\therefore \theta = 2.5 \times {10^{ - 4}}\,m$
So, the angular separation between the constructive bright fringes is $\theta = 2.5 \times {10^{ - 4}}\,m$.
Note: It should be remembered that, angular width is independent of the distance between the slits and the screen and the basic unit of conversions are like $1nm = {10^{ - 9}}m$ and a destructive interference is one where the intensity is low and thus forming a dark spot on the screen and also known as dark fringes. Constructive bright and destructive dark fringes are formed continuously one lead by another on the screen in Young’s double slit experiment.
Formula used:
Fringe width in Young’s double slit experiment is calculated as,
$\beta = \dfrac{D}{{\lambda d}}$
where, $D,d$ are the distance between slit and screen, and distance between slits and
$\lambda $ Is the wavelength of the wave.
Angular width formula is given by,
$\theta = \dfrac{\beta }{D}$
Complete step by step answer:
According to the question, we have given that,
$\lambda = 500\,nm = 500 \times {10^{ - 9}}\,m$ and $d = 2.0 \times {10^{ - 3}}\,m$
Now using the formula, $\theta = \dfrac{\beta }{D}$ we have,
$\theta = \dfrac{{D\lambda }}{{dD}}$
$\Rightarrow \theta = \dfrac{\lambda }{d}$
On putting the values we get,
$\theta = \dfrac{{500 \times {{10}^{ - 9}}}}{{2 \times {{10}^{ - 3}}}}$
$\therefore \theta = 2.5 \times {10^{ - 4}}\,m$
So, the angular separation between the constructive bright fringes is $\theta = 2.5 \times {10^{ - 4}}\,m$.
Note: It should be remembered that, angular width is independent of the distance between the slits and the screen and the basic unit of conversions are like $1nm = {10^{ - 9}}m$ and a destructive interference is one where the intensity is low and thus forming a dark spot on the screen and also known as dark fringes. Constructive bright and destructive dark fringes are formed continuously one lead by another on the screen in Young’s double slit experiment.
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