Question

A slit of width 0.53 mm is illuminated with light of wavelength 494 nm, and a screen is placed 116 cm in front of the slit. Find the widths of the first and second maxima on each side of the central maximum. w1= 1 mm (1st maxima) w2= 2 mm (2nd maxima)

Answer 1

The width of the first maximum is 2 mm and the width of the second maximum is 4 mm.

Step-by-step explanation:

The width of the central maximum in a single-slit diffraction pattern can be found using the formula:

w = (2 * (wavelength * distance)) / width

Where w is the width of the central maximum, wavelength is the wavelength of the light, distance is the distance from the slit to the screen, and width is the width of the slit. Plugging in the given values, we find:

w = (2 * (494 nm * 116 cm)) / 0.53 mm = 2 mm

Similarly, to find the width of the second maximum, we can use the formula:

w = (2 * (wavelength * distance)) / width

Using the given values, we find:

w = (2 * (494 nm * 116 cm)) / 0.53 mm = 4 mm