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Question

Monochromatic light of wavelength $$ 441 \mathrm{nm} $$ is incident on a narrow slit. On a screen $$ 2.00 \mathrm{m} $$ away, the distance between the second diffraction minimum and the central maximum is $$ 1.50 \mathrm{cm} $$. Find the width of the slit.


Solution

For the $$ m $$ th diffraction minimum, a $$ \sin \theta=m \lambda $$. 
We solve for the slit width:
$$a=\dfrac{m \lambda}{\sin \theta}=\dfrac{2(441 \mathrm{nm})}{\sin 0.430^{\circ}}=0.118 \mathrm{mm}$$

Physics
NCERT
Standard XII

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