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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.
Solution
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
\[y_2 - y_1 = \frac{D\left( y_2 - y_1 \right)}{d}\]
\[\Rightarrow y_2 - y_1 = \frac{1 . 5}{0 . 25 \times {10}^{- 3}}\left( 600 \times {10}^{- 9} - 480 \times {10}^{- 9} \right)\]
\[ y_2 - y_1 = 72 \times {10}^{- 5} m = 0 . 72 mm\]
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