Floaters. The floaters you see when viewing a bright, featureless background are diffraction patterns of defects in the vitreous humor that fills most of your eye. Sighting through a pinhole sharpens the diffraction pattern. If you also view a small circular dot, you can approximate the defect’s size. Assume that the defect diffracts light as a circular aperture does. Adjust the dot’s distance L from your eye (or eye lens) until the dot and the circle of the first minimum in the diffraction pattern appear to have the same size in your view. That is, until they have the same diameter $\mathit{D}\mathbf{\text{'}}$on the retina at distance $\mathit{L}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{\text{cm}}$from the front of the eye, as suggested in Fig. 36-42a, where the angles on the two sides of the eye lens are equal. Assume that the wavelength of visible light is $\mathit{\lambda }\mathbf{=}\mathbf{550}\mathbf{}\mathbf{\text{nm}}$. If the dot has diameter $\mathit{D}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{mm}}$and is distance $\mathit{L}\mathbf{=}\mathbf{45}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$from the eye and the defect is $\mathit{x}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{mm}}$ in front of the retina (Fig. 36-42b), what is the diameter of the defect?

Step by step solution

01

Concept/Significance of first minima in diffraction pattern

The angle corresponding to the first minima in diffraction pattern is given by,

$$\begin{gathered} \mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\frac{\mathbf{\lambda }}{\mathbf{d}} \\ \mathit{d}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{\lambda }}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }} \end{gathered}$$ ……. (1)

02

Find the diameter of the defect

From figure (a),

$$\begin{gathered} \frac{D\text{'}}{D}=\frac{L\text{'}}{L} \\ D\text{'}=D\left(\frac{L\text{'}}{L}\right) \\ =\left(2\text{mm}\right)\left(\frac{2\text{cm}}{45\text{cm}}\right) \\ =0.0889\text{mm} \end{gathered}$$

From figure (b),

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{{\displaystyle \raisebox{1ex}{$D^{\text{'}}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{x}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle \raisebox{1ex}{$0.0889mm$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{6.00mm}\right) \\ =0.424 \end{gathered}$$

Substitute 0.424 for $\theta$and $550\text{nm}$for $\lambda$in equation (1).

$$\begin{gathered} d=\frac{1.22\left(550\text{nm}\right)}{\sin \left(0.424\right)} \\ =\frac{1.22\left(550\times {10}^{-9}\text{m}\right)}{\sin \left(0.424\right)} \\ =9.067\times {10}^{-5}\text{m} \\ =91\mu m \end{gathered}$$

Therefore, the diameter of the defect is $91\mu m$.

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