Question

A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.

Answer 1

Given

Wavelengths of the source of light,

\[\lambda_1  = 480 \times  {10}^{- 9} m\text{ and }\lambda_2  = 600 \times  {10}^{- 9}  m\]

Separation between the slits,

\[d = 0 . 25  mm = 0 . 25 \times  {10}^{- 3}   m\]

Distance between screen and slit,

\[D = 150  cm = 1 . 5  m\]

We know that the position of the first maximum is given by

\[y = \frac{\lambda D}{d}\]

So, the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths = y₂ − y₁

\[y_2  -  y_1  = \frac{D\left( y_2 - y_1 \right)}{d}\]

\[\Rightarrow  y_2  -  y_1  = \frac{1 . 5}{0 . 25 \times {10}^{- 3}}\left( 600 \times {10}^{- 9} - 480 \times {10}^{- 9} \right)\]

\[           y_2  -  y_1  = 72 \times  {10}^{- 5}   m   = 0 . 72  mm\]