Question

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

Answer 1

Answer:

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Answer 2

Given that,

Wavelength = 590 nm

Wavelength = 596 nm

Aperture = 2 μm

Distance = 1.5 m

For 590 nm,

We need to calculate the separation between the positions of first maxima of the diffraction pattern

Using formula of separation

$x_{1}=\dfrac{(2n+1)}{2}\times\dfrac{\lambda D}{d}$

n = 1,

$x_{1}=\dfrac{3}{2}\times\dfrac{\lambda D}{d}$

Put the value into the formula

$x_{1}=\dfrac{3}{2}\times\dfrac{590\times10^{-9}\times1.5}{2\times10^{-6}}$

$x_{1}= 0.66375\ m$

For, 596 nm

$x_{2}=\dfrac{3}{2}\times\dfrac{\lambda D}{d}$

Put the value into the formula

$x_{2}=\dfrac{3}{2}\times\dfrac{596\times10^{-9}\times1.5}{2\times10^{-6}}$

$x_{2}=0.6705\ m$

We need to calculate the separation between the positions of the diffraction pattern

Using formula of separation

$x=x_{2}-x_{1}$

Put the value into the formula

$x=0.6705-0.66375$

$x=0.00675\ m$

Hence, The separation between the positions of first maxima of the diffraction pattern is 0.00675 m.