Question

Physics
Class XII
________________________________________________
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.
________________________________________________
$\sf \: @WildCat7083$​

Answer 1

Topic :-

Diffraction

Given :-

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.

To Find :-

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

Concept Used :-

Diffraction

When light passes through a narrow slit of width comparable to light's wavelength, the light flares out of the slit. This bending or spreading of light wave is called Diffraction.

$\sf {Position\:of\:maxima\:other\:than\:central\:maxima=\pm\left( n+\dfrac{1}{2}\right)\dfrac{\lambda D}{d}}$

where

$\sf {n =Order \:of\: maxima}$

$\sf{\lambda=Wavelength\:of\:light}$

$\sf{D=Distance\:between\:slit\:and\:screen}$

$\sf {d=Slit\:width}$

Solution :-

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.

$\sf {y_1=\left( 1+\dfrac{1}{2}\right)\dfrac{596\:nm\cdot D}{d}}$

$\sf {y_1=\left( \dfrac{3}{2}\right)\dfrac{596\:nm\cdot D}{d}}$

Similarly,

$\sf {y_2=\left( 1+\dfrac{1}{2}\right)\dfrac{590\:nm\cdot D}{d}}$

$\sf {y_2=\left( \dfrac{3}{2}\right)\dfrac{590\:nm\cdot D}{d}}$

(Note : First maxima of positive side is considered.)

Calculating the required seperation,

$\sf{y_1-y_2}$

$\sf {\left( \dfrac{3}{2}\right)\dfrac{596\:nm\cdot D}{d}-\left( \dfrac{3}{2}\right)\dfrac{590\:nm\cdot D}{d}}$

$\sf{\dfrac{3D}{2d}\left( 596\:nm-590\:nm \right)}$

$\sf{\dfrac{3D}{2d}\left( 6\:nm\right)}$

$\sf{\dfrac{18D}{2d}\:nm}$

$\sf{\dfrac{9D}{d}\:nm}$

Now, it is given that,

$\sf{D=1.5\:m}$

$\sf {d=2\times 10^{-6}\:m}$

Substituting values,

$\sf{\dfrac{9\times1.5\times 10^{-9}}{2\times 10^{-6}}\:m}$

$\sf {(\because 1\:nm=10^{-9}\:m)}$

$\sf{\dfrac{9\times 3\times 10^{6}}{2\times2\times 10^{9}}\:m}$

$\sf{\dfrac{27\times 10^{6-9}}{4}\:m}$

$\sf {6.75\times10^{-3}\:m}$

$\sf{6.75\:mm}$

$\sf {(\because 10^{-3}\:m=1\:mm)}$

Answer :-

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

Answer 2

Separation between two positions of first maxima .

$Δy=y2−y1=6.705−6.64=0.065mm</p><p></p><p>$