Answer to Question #256658 in Optics for Dion Gray

Question

An equiconvex lens having spherical surfaces of radius 10 cm, a central thickness of 2 cm, and a refractive index of 1.61 is situated between air and water (refractive index 1.33). An object 5 cm high is placed 60 cm in front of the lens surface. What will be the locations of the cardinal points for the lens (focal lengths in reference to the principal planes, principal and nodal points)?

What is the position of the image from the second principal plane and its nature?

Using matrix methods

Answer 1

1)

$\frac 1F=\frac{(n-1)d}{nR^2},$

$F=\frac{nR^2}{(n-1)d},$

$F=\frac{1.61\cdot 0.1^2}{0.61\cdot 0.02}=1.32~m,$

2)

$\frac{n_1}d+\frac{n_2}f=\frac 1F,\implies$

$f=\frac{n_2Fd}{n_1F-d},$

$f=\frac{1.32\cdot 0.6\cdot 1.33}{1\cdot 1.32-0.6}=1.46~m,$

$\Gamma=\frac fd=\frac Hh,\implies$

$H=h\frac fd,$

$H=0.05\cdot\frac{1.46}{0.6}=0.12~m.$