Question

light of wavelength 5000 A° is incident on a slit. the first minimum of the diffraction pattern is observed to lie at a distance of 5 mm from the central maximum on a screen placed at a distance of 3 m from the list. then the width of the list is​

Answer 1

Answer:

is sent a pic

handwritten

Explanation:

3 × 10^-4m

Answer 2

Answer:

The width of the slit is​ 0.3mm.

Explanation:

Given the wavelength of light, $5000 A° =5000\times 10^{-10} m$

Distance of the first minimum from the central maximum, $x=5 mm=5\times 10^{-3} m$

Distance between the screen and slit,  $D= 3 m$

Position of the nth minima is given by

$x_{n} =\frac{n\lambda D}{a}$

where a is the slit width, D is the distance between screen and slit and λ is the wavelength.

In the present case, as it is the first minimum, $n=1$

Substituting the given values, the slit width from the above formula is

$x_{1} =\frac{1\times \lambda D}{a}\implies a=\frac{\lambda D}{x_{1}}$

$\implies a =\frac{ 5000\times 10^{-10}\times 3}{5\times 10^{-3} }$

$= 1000\times 10^{-7}\times 3}$

$= 3\times 10^{-4}m=0.3mm$

Therefore, the slit width is​ 0.3mm.