Question

A plane wave of wavelength is incident on a slit with a width of $\mathbf{590}\mathbf{}\mathbf{nm}$. A thin converging lens of focal length $\mathbf{+}\mathbf{70}\mathbf{}\mathit{c}\mathit{m}$ is placed between the slit and a viewing screen and focuses the light on the screen. (a) How far is the screen from the lens? (b) What is the distance on the screen from the center of the diffraction pattern to the first minimum?

Short answer

(a) the distance between the screen and lens is 70 cm.

(b) the distance on the screen from the center of the diffraction pattern to the first minimum is $1.03\text{mm}$.

Step-by-step solution

Write the given data from the question.

The wavelength, $\lambda =590\text{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=5.90\times {10}^{-7}\mathrm{m}$

The slit width, $a=0.40\text{mm}\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=4.0\times {10}^{-4}\mathrm{m}$

The focal length of the lens, $f=+70\text{cm}\left(\frac{{10}^{-2}m}{1cm}\right)=0.70m$

Determine the formulas to calculate the distance of the screen and the distance between the screen and length.

The expression for the minima in the diffraction pattern is given as follows.

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$ $\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(i\right)$

Here, role="math" localid="1663092762632" $\mathit{a}$is the slit width, $\mathit{\lambda }$is the wavelength, role="math" localid="1663094164413" $\mathit{m}$is the number of fringes in the envelope, and role="math" localid="1663093303305" $\mathit{\theta }$is the angle of diffraction.

(a) Calculate the distance between the screen and the lens.

The light rays are the plane wave that causes to focus on the screen; the distance between the screen and the lens should equal the focal length of the lens.

Since the focal length of the lens is $+70\text{cm}$. Therefore, the distance between the screen and lens is$70\text{cm}$.

Hence the distance between the screen and lens is $70\mathrm{cm}$.

(b) Calculate the distance on the screen from the center of the diffraction pattern to the first minimum.

Consider the following figure shown below.

Here, $y$ is the distance to minima from central maxima.

The tangential angle is given by,

localid="1664270533266" $$\begin{gathered} \tan \theta =\frac{y}{D} \\ y=D\tan \theta \end{gathered}$$ (2)

Recall the equation (1),

localid="1664270544583" $\begin{array}{rcl}a\sin \theta & =& m\lambda \\ \sin \theta & =& \frac{m\lambda }{a}\\ \theta & =& {\sin }^{-1}\left(\frac{m\lambda }{a}\right)\end{array}$ (3)

Substitute $1$for $m$, $590\text{nm}$for localid="1664270604168" $\lambda$, and localid="1664270610711" $0.40\text{mm}$ for a into the above equation.

localid="1664270555244" $$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{1\times 5.90\times {10}^{-7}m}{4.00\times {10}^{-4}m}\right) \\ \theta ={\sin }^{-1}\left(1.475\times {10}^{-3}m\right) \\ \theta =0.0845° \end{gathered}$$

Substitute localid="1664270582863" $0.0845°$for localid="1664270590470" $\theta$and localid="1664270597193" $70\text{cm}$ for D into equation (2).

localid="1664270562634" $$\begin{gathered} y=0.70\text{m}\times \tan \left(0.0845°\right) \\ =0.70\text{m}\times 0.00147 \\ =0.00103\mathrm{m}\left(\frac{{10}^3\mathrm{mm}}{1\mathrm{m}}\right) \\ =1.03\text{mm} \end{gathered}$$

Hence the distance on the screen from the center of the diffraction pattern to the first minimum is localid="1664270575682" $1.03\text{mm}$.