91Ó°ÊÓ

1. The image size is 17.72 mm, The image is real, The image is inverted
2. The image is the same as part a with no difference at all

Lensmaker's equation for a thin lens:

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

f=The focal length of the lens, n = Index of refraction of lens material,$R_1$= radius of curvature of the first surface, $R_2$For the second surface.

Note: the two radii of curvature follow the signature rules.

$R_1,R_2$, is (+) when the radius in the back,(-) in front (the side of incident rays) f is (+) the lens is convergent, (-) the lens is divergent.

Sign rule for the radius of curvature of a surface of curvature (R): When the center of curvature C is on the same side as the outgoing light (the refracted light), the radius of curvature is positive, otherwise, it is negative

The lensmaker's equation gives us the unique focal length of each lens by using the index of refraction of the material and the two radii of curvature of the surfaces of the lens. There are two types of lenses diverging lens and a converging lens. Diverging lens causes the parallel incident rays to diverge while the converging lens causes the opposite.

Apply: in most problems, we use the lens maker equation to get the focal length of a lens. After that, we are asked to get the image or object position by using the equation for the thin lens.

Object -image relationship for ((thin lens))

$\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f}$

s=object distance from the lens, s'=The image distance from the lens, f=The focal length of the lens.

s is (+) in front of the lens, (-) in the back of the lens, s' is (+) in the back of the lens, (-) in front of the lens, f is (+) the lens is convergent, (-) the lens is divergent.

The sign rules for the variables in the equation:

The lens is reversed so the radius signs will reverse

1=the curved surface, 2=The flat surface, , ${\mathrm{R}}_2=∞,{\mathrm{R}}_1=+13\mathrm{cm}$

role="math" localid="1663915830148" $$\begin{gathered} \frac{1}{\mathrm{f}}=\left(\mathrm{n}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right) \\ \frac{1}{\mathrm{f}}=\left(1.7-1\right)\left(\frac{1}{∞}-\frac{1}{-13\mathrm{cm}}\right)=\frac{7}{130}\mathrm{cm}, \\ \mathrm{f}=\frac{130}{7}\mathrm{cm}=18.57\mathrm{cm} \\ \mathrm{s}\text{'}=\frac{\mathrm{fs}}{\mathrm{s}-\mathrm{f}}=\frac{\left(18.57\mathrm{cm}\right)\left(22.5\mathrm{cm}\right)}{\left(22.5\mathrm{cm}\right)-18.57\mathrm{cm}}=106.317\mathrm{cm} \end{gathered}$$

s' = 106.317 cm to the right of the lens (the back of the lens)

((The image is the same as part a with no difference at all)))