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# 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. - CBSE (Science) Class 12 - Physics

ConceptFraunhofer Diffraction Due to a Single Slit

#### 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.

#### Solution

For first maxima of the diffraction pattern we know  sintheta=(3lambda)/(2a)

where a is aperture of slit.

For small values of θ, sinθ tanθ = y/D

Where y is the distance of first minima from central line and D is the distance between the slit and the screen.

So

y= (3lambda)/(2a)D

For 590 nm,

y_1=(3xx590xx10^(-9))/(2xx2xx10^(-6))xx1.5

y1=0.66375 m

For 596 nm

y_2=(3xx596xx10^(-9))/(2xx2xx10^(-6))xx1.5

y2=0.6705 m

Separation between the positions of first maxima = y2− y1 = 0.00675 m

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Solution 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. Concept: Fraunhofer Diffraction Due to a Single Slit.
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