Question

Scientists shine a laser beam on a $35-\mu \mathrm{m}$-wide slit and produce a diffraction pattern on a screen $70\mathrm{cm}$behind the slit. Careful measurements show that the intensity first falls to $25\%$of maximum at a distance of$7.2\mathrm{mm}$from the center of the diffraction pattern. What is the wavelength of the laser light?

Hint: Use the trial-and-error technique demonstrated in Example $33.5$to solve the transcendental equation.

Short answer

Wavelength of the laser light is $600nm$.

Step-by-step solution

Calculation for intensity  

Intensity is ,

$I=I_0{\left[\frac{\sin (\pi ay/\lambda L)}{\pi ay/\lambda L}\right)}^2$

Where,

The maximum intensity is$I_0$,

Width of the slit is $a$,

The wavelength is $\lambda$

And the distance between the slit and the screen is $L$.

Rearrange the solution above.

localid="1651346221654" $\frac{I}{I_0}={\left[\frac{\sin (\pi ay/\lambda L)}{\pi ay/\lambda L}\right)}^2$

The intensity decreases to $25\%$of the high capacity$I_e$.

So,

$\frac{\sin (\pi ay/\lambda L)}{\pi ay/\lambda L}=\frac{1}{2}$

Calculation for sinxx

Use the trial-and-error method as outlined in the textbook's example.

Let$x=\pi ay/\lambda L$.

So,

$\frac{\sin x}{x}=\frac{1}{2}$

$=0.50$

The$x$is in radians.

The first minimum$y_1=\frac{\lambda L}{a}$

$x=\pi \mathrm{rad}$

This amount is lower than the solution. Because the intensity has tapered off more in this case, we should take our first forecast higher than the one given in the example.

For second trial, $x=1.9\mathrm{rad}$,

So,

$\frac{\sin x}{x}=\frac{\sin 1.9}{1.9}$

$=0.498$

For third trial, $x=1.89\mathrm{rad}$

So,

$\frac{\sin x}{x}=\frac{\sin 1.89}{1.89}$

role="math" localid="1651346909007" $=0.502$

Calculation for wavelength

Average of the value is,

$x=\frac{1.89+1.9}{2}$

$=1.895$

So,

$\frac{\sin x}{x}=0.5002$

Wavelength is,

$\lambda =\frac{\pi ay}{xL}$

$\lambda =\frac{\pi \left(35\mu \mathrm{m}\times \frac{{10}^{-6}\mathrm{m}}{1\mu \mathrm{m}}\right)\left(7.2\mathrm{mm}\times \frac{1\mathrm{m}}{1000\mathrm{mm}}\right)}{(1.895)\left(70\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right)}$

$=\left(6.0\times {10}^{-7}\mathrm{m}\right)\left(\frac{1\mathrm{nm}}{{10}^{-9}\mathrm{m}}\right)$

$=600nm$