Question

Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 × 10-6 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.​

Answer 1

• Let us consider position of first maxima formed by wavelength 596 nm as y₁.
• Let us consider position of first maxima formed by wavelength 590 nm as y₂.

$\sf {y_1=\left( 1+\dfrac{1}{2}\right)\dfrac{596\:nm\cdot D}{d}} \\ \\ \sf{y_1=\left( \dfrac{3}{2}\right)\dfrac{596\:nm\cdot D}{d}} \\ \\ \tt \: Similarly, \\ \sf {y_2=\left( 1+\dfrac{1}{2}\right)\dfrac{590\:nm\cdot D}{d}} \\ \\ \sf{y_2=\left( \dfrac{3}{2}\right)\dfrac{590\:nm\cdot D}{d}}$

Calculating the required separation

$\\\sf{y_1-y_2} \\ \\ \sf {\left( \dfrac{3}{2}\right)\dfrac{596\:nm\cdot D}{d}-\left( \dfrac{3}{2}\right)\dfrac{590\:nm\cdot D}{d}} \\ \\\sf{\dfrac{3D}{2d} \left( 596\:nm-590\:nm \right)} \\ \\ \sf{\dfrac{3D}{2d}\left( 6\:nm\right)} \\ \\ \sf{\dfrac{18D}{2d}\:nm} \\ \\ \sf{\dfrac{9D}{d}\:nm} \\ \\ \sf{D=1.5\:m} \\ \\ \sf {d=2\times 10^{-6}\:m} \\ \\ \tt \: Substituting \: values, \\ \sf{\dfrac{9\times1.5\times 10^{-9}}{2\times 10^{-6}}\:m} \\ \\ \sf {(\because \: 1\:nm=10^{-9}\:m)} \\ \sf{\dfrac{9\times 3\times 10^{6}}{2\times2\times 10^{9}}\:m} \\ \\ \sf{\dfrac{27\times 10^{6-9}}{4}\:m} \\ \\ \sf {6.75\times10^{-3}\:m}\sf{6.75\:mm}$

The separation between the positions of first maxima of the diffraction pattern obtained in the two cases is 6.75 mm.

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Answer 2

Explanation:

• separation between the positions of first maxima of the diffraction pattern obtained in the two cases is 6.75mm