Question

A beam of light consisting of two wavelengths, $6500\stackrel{\circ }{A}$ and $5200\stackrel{\circ }{A}$, is used to obtain interference fringes in a Young's double slit experiment. Find the distance of the third bright fringe on the screen from the central maximum for wavelength $6400\stackrel{\circ }{A}$.

• A. $1.17mm$ from the central fringe.
• B. $11mm$ from the central fringe.
• C. $1.17mm$ from the left fringe.
• D. $7mm$ from the central fringe.

Answer 1

The correct option is A $1.17mm$ from the central fringe.

$Given:$

$\text{Wavelength,}{\lambda }_1=6500A$

$\text{Wavelength,}{\lambda }_2=5200A$

$Solution:$

The distance of the mth bright fringe from the

central maximum is given by

$y_m=\frac{m\lambda D}{d}$

$y_3=\frac{3\lambda D}{d}$

$=\frac{3\times (6500\times {10}^{-10})\times 1.20}{2\times {10}^{-3}}$

$=1.17\times {10}^{-3}m=1.17mm$

$So,the$ $correct$ $option:A$