Question

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm. Calculate the wavelength of the light.

Answer 1

Solution:

Conditions for the position of the minima is,

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

Rearrange the above equations to get angle of diffraction,

$\theta=\tan ^{-1}\left(\frac{y}{D}\right)$

Substitute $1.35 \mathrm{~mm} for\ \mathrm{y} and\ 2.00 \mathrm{~m}\ for\ D.$

$\begin{aligned} \theta &=\tan ^{-1}\left(\frac{(1.35 \mathrm{~mm})\left(\frac{10^{-3} \mathrm{~m}}{1 \mathrm{~mm}}\right)}{2.00 \mathrm{~m}}\right) \\ &=0.038674^{\circ} \\ & \approx 0.0387^{\circ} \end{aligned}$

Expression for the wavelength of the light is,

$\lambda=\frac{d \sin \theta}{n}$

Substitute d for $0.750 \mathrm{~mm}, 0.0387^{\circ} for\ \theta, and\ 1 for\ n$ .

$\begin{aligned} \lambda &=\frac{(0.750 \mathrm{~mm})\left(\frac{10^{-3} \mathrm{~m}}{1 \mathrm{~mm}}\right) \sin \left(0.0387^{\circ}\right)}{1} \\ &=506.58 \times 10^{-9} \mathrm{~m} \\ & \approx 507 \mathrm{~nm} \end{aligned}$

The wavelength of the light is $507 \mathrm{~nm}.$