Answer to Question #287384 in Optics for Steven

Question

A small alien species with many tiny eyes is discovered. After analysis, we learn that light is focused on their transparent retina after being reflected off a concave mirror located at the back of their eyes. Assuming the eyes of this species only consist of a concave mirror with a radius of curvature of 420 µm immersed in a medium with a refractive index of 1.492:

 

a) Calculate the refractive power of the alien eye. (2 marks)

b) Where should the retina be placed to obtain the sharpest possible image for an object located 1.25 mm in front of the mirror? (3 marks)

 

Answer 1

Solution:

When immersed in a liquid radius of curvature of a mirror does not change.

a.

Given, radius of curvature of mirror R=420 micrometer.

Refractive index, n=1.492

Focal length

$\begin{aligned} &f=\frac{R}{2} \\ &f=\frac{420 \times 10^{-6}}{2} \\ &f=210 \times 10^{-6} \\ &f=210 \mu m \end{aligned}$

Refractive power

$\begin{aligned} &P=\frac{1}{f} \\ &\Rightarrow P=\frac{1}{210 \times 10^{-6}} \\ &\Rightarrow P=0.004762 \times 10^{6} \\ &\Rightarrow P=4 . 762 \times 10^{3} \text { dioptre } \end{aligned}$

b.

Object distance $u=-1.25 \mathrm{~mm}=-1.25\times 10^{-3}$ .

Focal-length $f=-210 \mu \mathrm{m}=-210\times10^{-6}$

From mirror formula

$\begin{aligned} &\frac{1}{f}=\frac{1}{V}+\frac{1}{u} \\ &\Rightarrow \frac{1}{V}=\frac{1}{f}-\frac{1}{u} \\ &\Rightarrow \frac{1}{V}=\frac{-1}{210 \times 10^{-6}}-\frac{1}{-1.25\times 10^{-3}} \\ &\Rightarrow \frac{1}{V}=-4 . 762 \times 10^{3}+0.8 \times 10^{3} \\ &\Rightarrow \frac{1}{V}=-3 . 962 \times 10^{3} \\ &\Rightarrow V=\frac{-1}{3. 962 \times 10^{3}} \\ &\Rightarrow V=-0 . 25 \times 10^{-3} \\ &\Rightarrow V=-0.25 \mathrm{~mm} \end{aligned}$