Question

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.

Answer 1

Answer:

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:

$y=\dfrac{L\lambda}{a}\\\\Where,y\to\ location\ of\ first\ minimum\ \textrm{diffraction}$

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, $W=2y=\frac{2L\lambda}{a}$

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

$W=\frac{2\times 2.40\ m\times 578\times 10^{-9}\ m}{0.0368\times 10^{-3}\ m}\\\\W=0.0754\ m$

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