In a single slit diffraction experiment, slit of width ' a ' is illuminated by light of wavelength ' λ ' and the width of the central maxima in diffraction pattern is measured as &

Question

In a single slit diffraction experiment, slit of width ' a ' is illuminated by light of wavelength ' λ ' and the width of the central maxima in diffraction pattern is measured as ' y '. When half of the slit is covered and illuminated by light of wavelength ( 1.5 ) λ , the width of the central maximum in diffraction pattern becomes

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The width of the central maximum in a single slit diffraction pattern is given by $y = \frac{2 λ D}{a}$ , where $D$ is the distance to the screen and $a$ is the slit width.

When half of the slit is covered, the new slit width becomes $a^{'} = \frac{a}{2}$ .

The new wavelength is given as $λ^{'} = 1.5 λ = \frac{3}{2} λ$ .

The new width of the central maximum is $y^{'} = \frac{2 λ ^{'} D}{a ^{'}}$ .

Substituting the values of $λ^{'}$ and $a^{'}$ , we get:

$y^{'} = \frac{2 ( \frac{3}{2} λ ) D}{\frac{a}{2}} = \frac{3 λ D}{\frac{a}{2}} = \frac{6 λ D}{a}$

Since $y = \frac{2 λ D}{a}$ , we can write $y^{'}$ in terms of $y$ :

$y^{'} = 3 (\frac{2 λ D}{a}) = 3 y$

Answer: 3 y

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