Question

A single slit that is 2100 nm wide forms a diffraction pattern when illuminated by monochromatic light of 680 nm wavelength. At an angle of 10 degrees from the central maximum, what is the ratio of the intensity to the intensity of the central maximum? Thanks..could anyone please show me the steps!

Answer 1

The Ratio of the intensity to the intensity of the central maximum is given by

$$ \begin{aligned} &\frac{I}{I_{\max }}=\left(\frac{\sin \beta}{\beta}\right)^{2} \\ &\text { where } \beta=\frac{\pi a \sin \theta}{\lambda} \\ &\beta=\frac{\pi \times 2100 \sin 10}{680}=1.6847 \mathrm{rad} \end{aligned} $$

in degrees

$$ \beta=1.6847 \times \frac{180}{\pi}=96.53^{\circ} $$

$$ \frac{I}{I_{\max }}=\left(\frac{\sin 96.53}{1.6847}\right)^{2}=0.3478 $$