A converging meniscus lens (see Fig.) with a refractive index of 1.52 has spherical surfaces whose radii are 7.00 cm and 4.00 cm. What is the position of the image if an object is placed 24.0 cm to the left of the lens? What is the magnification?

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, ${\mathrm{R}}_1$= radius of curvature of the first surface, ${\mathrm{R}}_2$For the second surface.

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

${\mathrm{R}}_1{\mathrm{R}}_2$,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.

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

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:

1. Sign rule for the object distance (s): when the object is on the same side of the refracting surface as the incoming light, object distance s is positive; otherwise, it is negative
2. Sign rule for the image distance (s dash): When the image is on the same side of the refracting surface as the outgoing light (the refracted light), the image distance is positive; otherwise, it is negative.

Note: the focal length depends on the curvature of the surfaces which forms the lens and depend on the material of the lens. Apply: in most problems, we are asked to get the position of the image forming from the refracted rays through a certain lens by using the focal length of the lens and the position of the object. In other problems, we are given the position of the object and the image to get the focal length of a lens.

Lateral magnification for a thin lens:

${\mathrm{R}}_1$=-7 cm negative as it is in the direction of incoming rays (in direction of the front of the lens) ${\mathrm{R}}_2=-4\mathrm{cm}$,

role="math" localid="1663917806446" $$\begin{gathered} \frac{1}{\mathrm{f}}=\left(\mathrm{n}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)=\left(1.52-1\right)\left(\frac{1}{-7\mathrm{cm}}-\frac{1}{-4\mathrm{cm}}\right)=\frac{39}{700} \\ \mathrm{f}=\frac{700}{39}=17.95\mathrm{cm} \\ \frac{1}{\mathrm{s}}+\frac{1}{\mathrm{s}\text{'}}=\frac{1}{\mathrm{f}} \\ \frac{1}{\mathrm{s}\text{'}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{s}}=\frac{\mathrm{s}-\mathrm{f}}{\mathrm{fs}} \\ \mathrm{s}\text{'}=\frac{\mathrm{fs}}{\mathrm{s}-\mathrm{f}}=\frac{\left(17.95\mathrm{cm}\right)}{\left(24\mathrm{cm}\right)-17.95\mathrm{cm}}=71.18\mathrm{cm} \end{gathered}$$

s' = 71.18 cm is positive as it is in the direction of the outcoming rays (the image is in the back of the lens)

$\mathrm{m}=\frac{-\mathrm{s}\text{'}}{\mathrm{s}}=\frac{\mathrm{y}\text{'}}{\mathrm{y}}$

m=The magnification, s=object distance, s'=The image distance, y'=The height of the image, y=The height of the object. m is (+) when the image is erect and(-) when the image is inverted.

$\mathrm{m}=\frac{-{\mathrm{s}}^{\text{'}}}{\mathrm{s}}=\frac{-71.18\mathrm{cm}}{24\mathrm{cm}}=-2.966$