How wide is the central diffraction peak on a screen 2.40 m behind a 0.0368-mm-wide slit illuminated by 578-nm light? Express your answer using three significant figures.

The width of the central diffraction peak is 0.0754 m.

Explanation:

Given:

Distance of screen from slit (L) = 2.40 m

Width of the slit (a) = 0.0368 mm = 0.0368 × 10⁻³ m

Wavelength of light (λ) = 578 nm = 578 × 10⁻⁹ m

Now, we know that, the location of the first diffraction minimum from the center is given as:

Now, in order to find the width of the central diffraction peak we need to consider the distance 'y' both above and below the central line.

Therefore, the width of the central diffraction peak is given as twice the distance 'y'.

So, Width of central diffraction peak is,

Now, plug in the values given and solve for width 'W'. This gives,

Therefore, the width of the central diffraction peak is 0.0754 m.

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

80cm/s.

24cm/s.

53.3cm/s.