Question

Monochromatic light of wavelength $\mathbf{441}\mathbf{}\mathbf{nm}$is incident on a narrow slit. On a screen $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$away, the distance between the second diffraction minimum and the central maximum is $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{cm}$. (a) Calculate the angle of diffraction role="math" localid="1663145961694" $\mathit{\theta }$of the second minimum. (b) Find the width of the slit.

Short answer

(a) The angle of diffraction for second order minima is $0.429°$.

(b) The width of slit is $1.18\times {10}^{-4}\text{m}$.

Step-by-step solution

Identification of given data

The wavelength of the monochromatic light is $\lambda =441\text{nm}$.

The distance of screen from narrow slit is $D=2\text{m}$.

The distance between central minima and second order minima is $w=1.50\text{cm}$.

The order of second minima is $m=2$.

Concept used

The diffraction is the phenomenon in which the light bends on the edges of the slit and pattern of light is formed on a screen.

Determination of angle of diffraction for second order minima

(a)

The angle of diffraction for second order minima is given as:

$\tan \theta =\frac{w}{D}$

Substitute all the values in equation.

$$\begin{gathered} \tan \theta =\frac{\left(1.50\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}{2\text{m}} \\ \theta =0.429° \end{gathered}$$

Therefore, the angle of diffraction for second order minima is $0.429°$.

Determination of width of slit

(b)

The width of slit is given as:

$m\lambda =d\sin \theta$

Substitute all the values in equation.

$$\begin{gathered} \left(2\right)\left(441\text{nm}\right)=d\left(\sin 0.429°\right) \\ d=1.18\times {10}^{-4}\text{m} \end{gathered}$$

Therefore, the width of slit is $1.18\times {10}^{-4}\text{m}$.