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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.
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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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