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?

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, $f=+70\text{cm}\left(\frac{{10}^{-2}m}{1cm}\right)=0.70m$

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.

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 $70\mathrm{cm}$.

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" $1.03\text{mm}$.