Refraction at a Spherical Surface

One Curved Interface, One Master Formula

Let light cross a single spherical surface of radius R from a medium of index $n_1$ into $n_2$ (object O on the axis, image I). For paraxial rays, the small-angle geometry plus Snell's law ($n_1 i = n_2 r$ for small angles) gives, after applying the Cartesian convention (OM $= -u$, MI $= +v$, MC $= +R$):

$$\boxed{\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}}$$

Reading the formula:

• It connects object and image across one refracting surface — the atomic unit from which the lens formulas are assembled (Section 6 applies it twice).
• R carries a sign: convex towards the incident light (centre beyond the surface) means $R > 0$.
• Set $R \to \infty$ (flat interface) and it reduces to $\frac{n_2}{v} = \frac{n_1}{u}$ — the apparent-depth result in disguise.

The Template Problem

Light from a point source in air falls on a glass surface (n = 1.5, R = 20 cm) from 100 cm away. Where is the image?

Setup: $n_1 = 1$, $n_2 = 1.5$, $u = -100$ cm, $R = +20$ cm.

$$\frac{1.5}{v} = \frac{1.5 - 1}{20} + \frac{1}{-100} = 0.025 - 0.010 = 0.015 \;\Rightarrow\; v = +100\text{ cm}$$

The image forms 100 cm inside the glass, in the direction of the incident light.

Problem-solving drill:

1. Identify $n_1$ (where light starts) and $n_2$ (where it goes) — the formula is direction-sensitive!
2. Sign u (almost always negative), sign R by where C lies.
3. Solve for v; positive v = image downstream (real, inside medium 2 here).

[JEE Tip] Curved-interface problems love two disguises: a fish under a curved bowl wall, and an air bubble inside a glass sphere (light then travels glass-to-air: $n_1 = 1.5$, $n_2 = 1$ — swap roles!). Direction of travel decides which index is which.

Solved Examples

Example 1: The NCERT template (Example 9.5)

A point source in air sits 100 cm from a convex glass surface (n = 1.5, R = 20 cm). Locate the image.

Solution:

1. $\frac{1.5}{v} - \frac{1}{-100} = \frac{0.5}{20}$.
2. $\frac{1.5}{v} = 0.025 - 0.01 = 0.015$.
3. $v = +100$ cm — a real image 100 cm beyond the surface, inside the glass.

Example 2: Same surface, nearer object [JEE Numerical]

Move the source to 25 cm from the same surface. Now where is the image?

Solution:

1. $\frac{1.5}{v} = \frac{0.5}{20} + \frac{1}{-25} = 0.025 - 0.04 = -0.015$.
2. $v = -100$ cm.
3. Negative v: a virtual image 100 cm on the incident side — close objects beat the surface's converging power.

Example 3: Flat-interface limit [NEET Numerical]

Use the master formula with $R \to \infty$ for a fish 90 cm under water (n = 4/3), viewed from straight above. Where does it appear?

Solution:

1. $\frac{n_2}{v} = \frac{n_1}{u}$ with $n_1 = 4/3$ (water, where light starts), $n_2 = 1$, $u = -90$.
2. $\frac{1}{v} = \frac{4/3}{-90}\cdot\frac{1}{1} \Rightarrow v = -67.5$ cm.
3. The fish appears 67.5 cm deep $= 90/(4/3)$ — the apparent-depth rule, straight from the master formula.

Example 4: Glass-to-air crossing [JEE Numerical]

An object inside glass (n = 1.5) sits 30 cm from a surface separating glass from air; the surface is concave as seen from inside the glass, with signed radius R = -10 cm. Find the image. (Light goes glass to air: $n_1 = 1.5$, $n_2 = 1$.)

Solution:

1. $\frac{1}{v} - \frac{1.5}{-30} = \frac{1 - 1.5}{-10} = 0.05$.
2. $\frac{1}{v} = 0.05 - 0.05 = 0$.
3. $v \to \infty$: the rays emerge parallel — this object happens to sit at the surface's first focus.

Example 5: Sign the radius [NEET pattern]

State the sign of R when light strikes (a) a surface bulging towards it (convex), (b) a surface curving away (concave).

Solution:

1. R runs from the surface's pole to its centre of curvature C, signed along the incident light.
2. (a) Convex towards the light: C lies downstream — $R > 0$.
3. (b) Concave towards the light: C lies upstream — $R < 0$.

Example 6: The formula's direction-sensitivity

Why must $n_1$ always be the medium the light comes FROM?

Solution:

1. The derivation applies Snell's law at the crossing — $n_1\sin i = n_2\sin r$ with i in the first medium.
2. Swapping the labels flips the sign of $(n_2 - n_1)$ and misplaces the image.
3. In bubble-in-glass problems light starts in glass: $n_1 = 1.5$, $n_2 = 1$. Read the ray, then label.

Example 7: Landing on the second focus [JEE Numerical]

For the air-glass surface (n = 1.5, R = +20 cm), find the object distance for which the image forms at $v = +60$ cm.

Solution:

1. $\frac{1.5}{60} - \frac{1}{u} = \frac{0.5}{20} = 0.025$.
2. But $\frac{1.5}{60} = 0.025$ exactly, so $\frac{1}{u} = 0$, i.e. $u \to \infty$.
3. A parallel beam (object at infinity) focuses at 60 cm — incidentally the surface's second focal distance: $f_2 = \frac{n_2 R}{n_2 - n_1} = \frac{1.5 \times 20}{0.5} = 60$ cm.

Example 8: First and second foci of one surface [JEE Numerical]

For the same surface, find the first focal distance (object position giving parallel emergent rays).

Solution:

1. Set $v \to \infty$: $-\frac{n_1}{u} = \frac{n_2 - n_1}{R}$.
2. $u = -\frac{n_1 R}{n_2 - n_1} = -\frac{1 \times 20}{0.5} = -40$ cm.
3. Note $|f_1| \ne |f_2|$ (40 vs 60 cm) — single surfaces have unequal foci, in the ratio $n_1 : n_2$.

Example 9: A drop as a surface [NEET Numerical]

A tiny insect sits at the centre of curvature of a hemispherical dew drop (n = 4/3, R = 3 mm) and is viewed from above through the curved surface. Where does it appear?

Solution:

1. An object at the centre of curvature sends every ray along a radius — each meets the surface normally and passes undeviated.
2. The formula agrees: with the signed radius $-R$ (C lies on the incident side) and $u = -R$: $\frac{n_2}{v} = \frac{n_2-n_1}{-R} + \frac{n_1}{-R} = \frac{n_2}{-R}$, so $v = -R$.
3. The insect appears exactly where it is — rays through C never bend. A favourite shortcut.

Example 10: Building towards the lens

How does this section's formula become a lens formula?

Solution:

1. A thin lens is two spherical surfaces back-to-back.
2. Apply $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R}$ at the first surface; its image becomes the object for the second application.
3. Add the two equations, let the thickness vanish — out drops the lens maker's formula (Section 6). One brick, used twice.