Question

In Young's double slit experiment, the fringes are formed at a distance of $1\text{m}$ from double slit of separation $0.12\text{mm}$. Given $\lambda =6000\stackrel{\circ }{A}$. Then the distances of 3rd dark band from the centre of the screen and 3rd bright band from the centre of the screen will be respectively

• A. $1.25\text{cm},1.5\text{cm}$
• B. $2.25\text{cm},2.5\text{cm}$
• C. $3.25\text{cm},3.5\text{cm}$
• D. $0.25\text{cm},0.5\text{cm}$

Answer 1

The correct option is A $1.25\text{cm},1.5\text{cm}$


Distance of $n^{th}$ bright band from the centre is given by
$y=\frac{n\lambda D}{d}$
Distance of $n^{th}$ dark band from the centre is given by
$y=\left(\frac{2n-1}{2}\right)\frac{\lambda D}{d}$

And the pattern will look like


Substituting the values $D=1m,d=0.12\times {10}^{-3}m$ and $\lambda =6000\times {10}^{-10}m$ in the two equations, we get
Distance of 3rd dark band $y_3=1.25cm$ and
Distance of 3rd bright band $y_3=1.5cm$ and