Q44E - University Physics with Modern Physics

Question

The Lens of the Eye: The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 mm, which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object 16 cm tall is placed 30.0 cm from the eye lens, where would the lens focus it and how tall would the image be? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because the lens is embedded in fluids having refractive indexes different from that of air.)

Step-by-Step Solution

Answer

a) the radii of curvature of its two surfaces are +7.04 mm and -7.04 mm.

b) the image formed by the lens is 0.822 cm to the right of the lens having a height of 0.438 cm and it is real and inverted in nature with respect to the object.

Step 1: Basic definition

An optical lens is a transparent transmissive optical component that is used to either converge or diverge the light emitting from a object. These transmitted light forms an image of that object.

Thin Lens formula:

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

Here, 

u = Object distance from lens

v = Image distance from lens

f = Focal length of the lens

Sign Convention:

1. Object Distance (u): (+) in front of lens; (-) in the back of lens
2. Image Distance (v): (-) in front of lens; (+) in the back of lens
3. Focal Length (f): (+) for convergent (convex) lens; (-) for divergent (concave) lens

Lensmaker’s equation for a thin lens:

$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R^2}-\frac{1}{R^2}\right)$

Here,

n = Refractive index of the lens material

R₁ = Radius of curvature of first surface

R₂ = Radius of curvature of second surface

Sign Convention:

1. If R₁ & R₂ are positive – Radius in the back
2. If R₁ & R₂ are negative – Radius in the front

Lateral magnification (m): It is defined as the ratio of the height of image to the height of the object viewed through a lens. It is a dimensionless quantity.

 $\mathit{m}\mathbf{=}\mathbf{-}\frac{\mathbf{v}}{\mathbf{u}}\mathbf{=}\frac{{\mathbf{H}}_{\mathbf{i}}}{{\mathbf{H}}_{\mathbf{0}}}$

Here,

H_o = Height of the object, and H_i = Height of the image

Sign Convention:

1. If m is positive – Image is erected
2. If m is negative – Image is inverted

Step 2: Radii of curvature of lens

For a double convex lens: 

By using Lensmaker’s formula:

$$\begin{gathered} \frac{1}{f}=\left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \frac{1}{8mm}=\left(1.44-1\right)\left(\frac{1}{R_1}-\frac{1}{-R_2}\right) \\ \frac{1}{8mm}=\left(1.44-1\right)\left(\frac{2}{R_1}\right) \\ {R}_1=2*0.44*8 \\ {R}_1=7.04mm \\ {R}_2=-R_1=-7.04mm \end{gathered}$$ 

Therefore, the radii of curvature of its two surfaces are +7.04 mm and -7.04 mm.

Step 3: Location and Nature of Image

Focal length of 1^st lens, f₁ = +8 mm = +0.8 cm

Height of object, H_o = 16 cm

Object distance, u = +30.0 cm 

 

By using lens formula

 

$$\begin{gathered} \Rightarrow \frac{1}{u}+\frac{1}{v}=\frac{1}{f_1} \\ \Rightarrow \frac{1}{+30}+\frac{1}{v_1}=\frac{1}{+0.8} \\ \Rightarrow \frac{1}{v_1}=\frac{1}{+0.8}-\frac{1}{+30} \\ \Rightarrow \frac{1}{v_1}=\frac{30-0.8}{30*0.8} \\ \Rightarrow v_1=+0.822cm \end{gathered}$$

 

Lateral magnification for a lens is given by

 

$m=-\frac{v}{u}=\frac{H_i}{H_0}$

 

Thus, height of the image is given by

 

$$\begin{gathered} H_i=\left(-\frac{v_1}{u}\right)H_0 \\ =\left(\frac{0.822cm}{30cm}\right)*16cm \\ =-0.438cm \end{gathered}$$

 

Therefore, the image formed by the lens is 0.822 cm to the right of the lens having a height of 0.438 cm and it is inverted in nature with respect to the object.