Question

An insect 3.75 mm tall is placed 22.5 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 13.0 cm, and the index of refraction of the lens material is 1.70.
(a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted?
(b) Repeat part (a) if the lens is reversed.

Answer 1

Solution :

a). The focal length of the lens is given as :

$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f}=(1.70-1)\left(\frac{1}{\infty}-\frac{1}{-13}\right)$

$f=18.57 \ cm$

The thin lens equation is :

$\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$

The distance of the image from the lens is

$d_i=\left(\frac{1}{f}-\frac{1}{d_o}\right)^{-1}$

$d_i=\left(\frac{1}{18.57}-\frac{1}{22.5}\right)^{-1}$

  $=1.06 \ m$

The magnification of the lens is :

$m=-\frac{d_i}{d_o}=\frac{h_i}{h_o}$

The height of the image is :

$h_i=-\frac{d_i}{d_o}(h_o)$

$h_i=-\frac{106.3 \ cm}{22.5 \ cm}(3.75 \ mm)$

   $=-17.7 \ cm$

Thus, the real and the inverted image is formed at a distance of 1.06 m. And the size of the image is 17.7 cm

b). When the lens is reversed.

The focal length of the lens is given as :

$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f}=(1.70-1)\left(\frac{1}{13}-\frac{1}{\infty}\right)$

$f=18.57 \ cm$

The thin lens equation is :

$\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$

The distance of the image from the lens is

$d_i=\left(\frac{1}{f}-\frac{1}{d_o}\right)^{-1}$

$d_i=\left(\frac{1}{18.57}-\frac{1}{22.5}\right)^{-1}$

  $=1.06 \ m$

The magnification of the lens is :

$m=-\frac{d_i}{d_o}=\frac{h_i}{h_o}$

The height of the image is :

$h_i=-\frac{d_i}{d_o}(h_o)$

$h_i=-\frac{106.3 \ cm}{22.5 \ cm}(3.75 \ mm)$

   $=-17.7 \ cm$

Thus, when the lens is reversed, a real and an inverted image is formed at a distance of 1.06 m. And the size of the image is 17.7 cm