Answer to Question #285644 in Optics for NICKO

Question #285644

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.

1
Expert's answer
2022-01-12T08:31:43-0500

Solution:

Conditions for the position of the minima is,

tanθ=yD\tan \theta=\frac{y}{D}

Rearrange the above equations to get angle of diffraction,

θ=tan1(yD)\theta=\tan ^{-1}\left(\frac{y}{D}\right)

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

θ=tan1((1.35 mm)(103 m1 mm)2.00 m)=0.0386740.0387\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,

λ=dsinθn\lambda=\frac{d \sin \theta}{n}

Substitute d for 0.750 mm,0.0387for θ,and 1for n0.750 \mathrm{~mm}, 0.0387^{\circ} for\ \theta, and\ 1 for\ n .

λ=(0.750 mm)(103 m1 mm)sin(0.0387)1=506.58×109 m507 nm\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 nm.507 \mathrm{~nm}.


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