Question

Distance of ${{5}^{th}}$ dark fringe from centers is 4 mm. If D=2 m, $\lambda =600nm$, then distance between slits is
A. 1.35 mm
B. 2.00 mm
C. 2.35 mm
D. 10.35 mm

Answer 1

Young’s double slit experiment is performed in the question and we are given the distance of the ${{5}^{th}}$ dark fringe from centre. We are also given the wavelength and the distance between slit and screen. By using the equation to find the distance of ${{n}^{th}}$ dark fringe and the equation for fringe width, we will get the required solution.

Formula used:
${{n}^{th}}\text{fringe distance}=\dfrac{\left( 2n-1 \right)\beta }{2}$
$\beta =\dfrac{\lambda D}{d}$

Complete step by step answer:
In the question a Young’s double slit experiment is performed and we are given the distance of the ${{5}^{th}}$ dark fringe from the centre as 4 mm.
Converting this into meters, we will get the distance of the ${{5}^{th}}$ dark fringe from the centre will be $4\times {{10}^{-3}}m$.
We are also given the distance between the slit and the screen as, $D=2m$ and the wavelength of the light used as $\lambda =600nm$.
Converting the value of wavelength from nanometer to meter, we will get
$\Rightarrow \lambda =600\times {{10}^{-9}}m$
We know that the distance of the ${{n}^{th}}$ dark fringe from the centre is given by the equation,
$\dfrac{\left( 2n-1 \right)\beta }{2}$, where ‘$\beta$’ is the fringe width.
Then for the ${{5}^{th}}$ dark fringe, we can write this equation as,
$\Rightarrow \dfrac{\left( 2\times 5-1 \right)\beta }{2}$
By solving this we will get,
$\Rightarrow \dfrac{\left( 10-1 \right)\beta }{2}$
$\therefore \dfrac{9\beta }{2}$
We also know that the fringe width, ‘$\beta$’ is given by the expression,
$\beta =\dfrac{\lambda D}{d}$, were ‘$\lambda$’ is the wavelength of the light used, ‘D’ is the distance between the slit and the screen and ‘d’ is the distance between the two slits.
By substituting the value of ‘$\beta$’ in the equation for the distance of ${{5}^{th}}$ dark fringe, we will get
$\Rightarrow \dfrac{9}{2}\times \dfrac{\lambda D}{d}$
In the question we are given the distance of the ${{5}^{th}}$ dark fringe from the centre. Therefore we can equate the given value with the above equation. Thus we will get,
$\Rightarrow 4\times {{10}^{-3}}=\dfrac{9\lambda D}{2d}$
We are asked to find the distance between two slits, i.e. ‘d’. from the above equation we can write ‘d’ as,
$\Rightarrow d=\dfrac{9\lambda D}{2\left( 4\times {{10}^{-3}} \right)}$
By substituting the value for wavelength and the distance between slit and screen in the above equation we will get,
$\Rightarrow d=\dfrac{9\times 600\times {{10}^{-9}}\times 2}{2\left( 4\times {{10}^{-3}} \right)}$
By solving this we will get,
$\Rightarrow d=\dfrac{9\times 600\times {{10}^{-6}}}{4}$
$\Rightarrow d=9\times 150\times {{10}^{-6}}$
$\Rightarrow d=1.35\times {{10}^{-3}}m$
$\therefore d=1.35mm$
Therefore the distance between the two slits is 1.35 mm.
Hence the correct answer is option A.

Note:
Young’s double slit experiment is done to prove the wave nature of light. It also demonstrates the constructive and destructive interference of light. In the experiment two parallel slits are made and the light from the source is made to pass through the slits and fall on the screen.
Constructive interference takes place when the path difference of the two waves is the integral multiple of the wavelength of the light and destructive interference occurs when the path difference is half – integral multiple of the wavelength.