Question

How wide is the central diffraction peak on a screen 2.20 m behind a 0.0348 mm wide slit illuminated by419 nm light?
_______ cm

Answer 1

$$ \begin{aligned} &\lambda=419 \mathrm{~nm}=419 \times 10^{-9} \mathrm{~m} \\ &D=0.0348 \mathrm{~mm}=0.0348 \times 10^{-3} \mathrm{~m} \end{aligned} $$

Now we know the formula \(\sin \theta=\frac{\mathrm{m} \lambda}{\mathrm{D}}=\frac{(1)\left(419 \times 10^{-9} \mathrm{~m}\right)}{0.0348 \times 10^{-3} \mathrm{~m}}\) $$ \theta=0.69^{\circ} $$

$$ \text { Nowy } \begin{aligned} y_{1} &=L \tan \theta_{1} \\ &=(2.20 \mathrm{~m}) \tan 0.69^{\circ} \\ &=0.026 \mathrm{~m} \end{aligned} $$

Now width of the central diffraction peak is \(\Delta y=2 y=2(0.026 \mathrm{~m})=0.053 \mathrm{~m}\)