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A Source Emitting Light of Wavelengths 480 Nm and 600 Nm is Used in a Double-slit Interference Experiment. - Physics

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प्रश्न

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.

योग

उत्तर

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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अध्याय 17: Light Waves - Exercise [पृष्ठ ३८१]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 17 Light Waves
Exercise | Q 10 | पृष्ठ ३८१

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