93.5k views
3 votes
A single slit of width 0.2 mm is illuminated by a mercury light of wavelength 576 nm. Find the intensity at a 20° angle to the axis in terms of the intensity of the central maximum?

1 Answer

4 votes

Final answer:

To find the intensity at a 20° angle in a single-slit diffraction pattern, one must use the diffraction formula to calculate beta, and then use it to find the relative intensity compared to the central maximum.

Step-by-step explanation:

The question involves calculating the intensity of light at a certain angle from the central maximum in a single-slit diffraction pattern. The intensity can be calculated using the formula of diffraction minima for a single slit, which is a sin(\(\theta\)) = m\(\lambda\), where a is the slit width, \(\theta\) is the diffraction angle, m is the order number, and \(\lambda\) is the wavelength of the light. However, to find the relative intensity, one would use the formula \(I(\theta) = I_0 \left(\frac{\sin(\beta)}{\beta}\right)^2, with \(\beta = \frac{\pi a \sin(\theta)}{\lambda}. In the case of a slit of width 0.2 mm illuminated by mercury light with a wavelength of 576 nm at an angle of 20°, first, we need to calculate \(\beta and then use it to determine the relative intensity compared to that of the central maximum, denoted by I_0.

User Emad Gabriel
by
9.2k points