Q35E - University Physics with Modern Physics

Question

The cornea behaves as a thin lens of focal length approximately1.8 cm, although this varies a bit. The material of which it is made has an index of refraction of 1.38 cm , and its front surface is convex, with a radius of curvature of 5.0 mm. (a) If this focal length is in air, what is the radius of curvature of the back side of the cornea? (b) The closest distance at which a typical person can focus on an object (called the near point) is about 25cm, although this varies considerably with age. Where would the cornea focus the image of an 8.00 mm -tall object at the near point? (c) What is the height of the image in part (b)? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because, on one side, the cornea has a fluid with a refractive index different from that of air.)

Step-by-Step Solution

Answer

1. The radius of curvature of the backside of the cornea is 18.6 mm
2. The distance where the cornea will focus the image of an 8.0- mm -tall object at the near point is 19 mm 
3. The height of the image formed in case (b) is -0.61mm

Step 1: (a) Determination of the radius of curvature of the backside of the cornea.

The lens maker’s formula is,

$\frac{1}{\mathrm{f}}=\left(\mathrm{n}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)$  

Here, f is the focal length, n is the refractive index and the R’s are the radii of curvature.

Solve for the radius of curvature R₂ when R₁ = +5.0 mm

 $$\begin{gathered} \frac{1}{\mathrm{f}\left(\mathrm{n}-1\right)}=\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2} \\ \frac{1}{{\mathrm{R}}_2}=\frac{1}{{\mathrm{R}}_1}-\frac{1}{\mathrm{f}(\mathrm{n}-1)} \\ =\frac{1}{+0.5 \mathrm{mm}}-\frac{1}{\left(18.0 \mathrm{mm}\right)\left(0.38\right)} \\ =18.6 \mathrm{mm} \end{gathered}$$

Thus, the radius of curvature R₂ is 18.6 mm. 

Step 2: (b) Determination of the distance where the cornea will focus the image of an 8.0-mm-tall object at the near point.

The expression in this case connecting the image distance v  the object distance u and the focal length f is,

$$\begin{gathered} \frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}} \\ \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ =\frac{\mathrm{u}-\mathrm{f}}{\mathrm{uf}} \end{gathered}$$                

So,

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}} \\ =\frac{\left(25 \mathrm{cm}\right)\left(1.8 \mathrm{cm}\right)}{25 \mathrm{cm}-1.8\mathrm{cm}} \\ =1.9 \mathrm{cm} \\ =19 \mathrm{mm} \end{gathered}$$

Thus, the image location is   and it is certainly not at the retina.       

Step 3: (b) Determination of the height of the image formed in case (b).

Magnification of a lens is nothing but the lens’s capability to magnify the image of any regular size object. Mathematically,

  $$\begin{gathered} \mathrm{m}=\frac{\mathrm{y}\text{'}}{\mathrm{y}} \\ =-\frac{\mathrm{v}}{\mathrm{u}} \\ \therefore \mathrm{m}=-\frac{1.9\mathrm{cm}}{25 \mathrm{cm}} \\ =-0.076 \\ \mathrm{y}\text{'}=\mathrm{my} \\ =\left(-0.076\right)\left(8.0 \mathrm{mm}\right) \\ =-0.61\mathrm{mm} \end{gathered}$$

So, height of the image is -0.61mm .