Question

White light is passed through a double slit and interference is observed on a screen $1.5\text{m}$ away. The separation between the slits is $0.3\text{mm}$. The first violet and red fringes are formed $2.0\text{mm}$ and $3.5\text{mm}$ away from the central white fringe. The difference in wavelengths of red and violet light is nm

• A. 300
• B. 300.0

Answer 1

Answer: (A), (B)

Given :-
$D=1.5\text{m}$
$d=0.3\text{mm}$
$(y_1{)}_V=2.0\text{mm}$
$(y_1{)}_R=3.5\text{mm}$

The distance of $n$ the bright fringe from the center is :
$y=\frac{n\lambda D}{d}$
For first bright fringe $n=1$
$\Rightarrow y_1=\frac{\lambda D}{d}$

For violet light :
$\Rightarrow 2\times {10}^{-3}=\frac{{\lambda }_V\times 1.5}{0.3\times {10}^{-3}}$
$\Rightarrow {\lambda }_V=4\times {10}^{-7}\text{m}$

For red light :
$\Rightarrow 3.5\times {10}^{-3}=\frac{{\lambda }_V\times 1.5}{0.3\times {10}^{-3}}$
$\Rightarrow {\lambda }_R=7\times {10}^{-7}\text{m}$

Therefore,
${\lambda }_R-{\lambda }_V=3\times {10}^{-7}\text{m}$
${\lambda }_R-{\lambda }_V=300\text{nm}$