Physics · Ch 6 — Optics

Equation for Refraction at Single Spherical Surface

6.5.1

Equation for Refraction at Single Spherical Surface

For two media of refractive index $n_1$ (containing a point object $O$ ) and $n_2$ , separated by a spherical surface of radius of curvature $R$ and centre of curvature $C$ , a paraxial ray from $O$ strikes the surface at $N$ and bends towards the normal $NC$ (since $n_2>n_1$ ), crossing the axis at the image point $I$ . Using Snell's law in the small-angle form $n_1 i = n_2 r$ , and the small angles $\alpha=\angle NOP$ , $\beta=\angle NCP$ , $\gamma=\angle NIP$ that satisfy $i=\alpha+\beta$ (triangle $ONC$ ) and $r=\beta-\gamma$ (triangle $INC$ ), then substituting the small-angle approximations $\alpha=PN/PO$ , $\beta=PN/PC$ , $\gamma=PN/PI$ and applying the Cartesian sign convention ( $PO=-u$ , $PI=+v$ , $PC=+R$ ), the algebra simplifies down to $\boxed{\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfra …

Figure 6.31Refraction at single spherical surface

What this figure shows. A point object O sits in a medium of refractive index n1, to the left of a spherical interface of radius of curvature R and centre of curvature C, beyond which lies a second medium of refractive index n2 (with n2 greater than n1). A paraxial ray from O strikes the surface at N and bends towards the normal NC, crossing the axis at the image point I inside the denser medium. The small angles this ray, the radius NC, and the refracted ray each make with the axis (labelled alpha, beta, gamma in the derivation) are exactly what the equation n2/v - n1/u = (n2-n1)/R is built from …

Misc Example 6.12Locating the image formed by a single curved surface separating air and glass

Worked out. An object sits 15 cm to the left (u = -15 cm) of a spherical surface of radius of curvature R = 30 cm, separating air (n1 = 1) from glass (n2 = 1.5). Substituting into n2/v - n1/u = (n2-n1)/R gives 1.5/v - 1/(-15) = (1.5-1)/30 = 1/60, so 1.5/v = 1/60 - 1/15 = -1/20, giving v = -30 cm. The negative sign shows the image forms 30 cm to the left of the surface, i.e. it is a virtual image, on the same side as the object, formed inside the rarer medium rather than the denser one. …

Misc Example 6.13Size of the image formed at a curved air-water interface

Worked out. An object of height 1.0 cm sits 40 cm from a spherical surface (R = -20 cm) separating air (n1 = 1) from water (n2 = 1.33). Substituting into n2/v - n1/u = (n2-n1)/R with u = -40 cm gives 1.33/v - 1/(-40) = (1.33-1)/(-20) = -0.0165, and solving gives v = -32.0 cm, a virtual image 32.0 cm to the left of the surface. The magnification is m = (n1 v)/(n2 u) = (1 times -32)/(1.33 times -40) = 0.6, so the image height is 1.0 times 0.6 = 0.6 cm; since m is positive the image is erect (and, as located, virtual) -- a smaller, upright image formed on the …

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