Question

A beam of light containing two wavelengths 5200 $\mathring{A}$ and 6500 $\mathring{A}$ is used in Young's experiment to obtain interference fringes. What is the least distance from the central maximum on the screen where the bright fringes due to both wavelengths coincide? (Given the ance between slits is 2 mm and distance of the en from slits is 120 mm.)

• A. $0.156mm$
• B. $0.312mm$
• C. $0.78mm$
• D. $1.1mm$

Answer 1

The correct option is A $0.156mm$

We have

$y=\frac{n\lambda D}{d}$

where $y$ is the distance from central maximum to the centre of the $n^{th}$ bright fringe

If $m^{th}$ bright fringe of $6500A^o$ coincides with the $n^{th}$ bright fringe of $5200A^o$ then

$\frac{m\times 6500\times D}{d}=\frac{n\times 5200\times D}{d}$

$⟹\frac{m}{n}=\frac{4}{5}$

Hence the minimum value of $m$ and $n$ that satisfy this equation are $4$ and $5$.

Distance of $4^{th}$ bright fringe of $6500A^o$ or $5^{th}$ bright fringe of $5200A^o$ from central maximum

$\therefore y=\frac{4\times 6500\times {10}^{-10}\times 0.12}{2\times {10}^{-3}}=0.156mm$