In double slit experiment, the wavelength of the light source 400… - Vidyadip

Question

In double slit experiment, the wavelength $\lambda $ of the light source $400\ nm$ , the slit seperation $d$ is $20\ \mu m$ and the individual slit width a is $4\ \mu m$ . Consider the interference of the light from the two slits and also the diffraction of light through each slit. Find number bright interference fringes are within the central peak of diffraction envelope

✓

Answer

d
$\mathrm{d} \sin \theta=\mathrm{n} \lambda$ for interference

$\mathrm{d} \sin \theta=\lambda$ for diffraction

$\frac{d}{a}=n=\frac{20}{4}=5$

so interference maximas is in $9$

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