Light – Reflection and Refraction • Topic 3 of 3

Refraction & Lenses

When light passes from one transparent medium into another it changes speed, and as a result it bends. This bending of light at the boundary of two media is called refraction. Light bends because its speed is different in different media. Going from a rarer medium (such as air) into a denser one (such as glass or water), light slows down and bends towards the normal; going from denser to rarer it speeds up and bends away from the normal. A ray striking the surface along the normal (at $0^\circ$) passes straight through without bending.

Refraction obeys two laws. The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane. The second law is Snell's law: for a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant, $\frac{\sin i}{\sin r} = n$, where $n$ is the refractive index of the second medium with respect to the first. The refractive index can also be written in terms of speeds:

$n = \frac{c}{v}$, where $c$ is the speed of light in vacuum (about $3 \times 10^8\,\text{m/s}$) and $v$ is the speed of light in the medium.

A lens is a piece of transparent material bounded by two curved surfaces. A convex (converging) lens is thicker in the middle and brings parallel rays to a real focus; a concave (diverging) lens is thinner in the middle and spreads parallel rays so they appear to come from a focus. Each lens has two foci, one on each side, and an optical centre O through which a ray passes undeviated.

The ray rules for lenses are straightforward. A ray parallel to the principal axis passes through the focus (convex) or appears to diverge from the focus (concave) after refraction; a ray through the optical centre goes straight on. Distances follow the same New Cartesian convention as for mirrors. The lens formula is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ — note the minus sign, unlike the mirror formula. Magnification for a lens is $m = \frac{h'}{h} = \frac{v}{u}$. A convex lens has a positive focal length and a concave lens a negative one. The power of a lens measures how strongly it converges or diverges light: $P = \frac{1}{f}$, where $f$ is in metres and the unit of power is the dioptre (D). A convex lens has positive power and a concave lens has negative power.

Solved Examples

Worked Example

The speed of light in glass is $2 \times 10^8\,\text{m/s}$ and in vacuum it is $3 \times 10^8\,\text{m/s}$. Find the refractive index of glass.

Solution

Refractive index $n = \frac{c}{v}$.
$n = \frac{3 \times 10^8}{2 \times 10^8}$.
$n = 1.5$.

Answer: Refractive index of glass $= 1.5$.

Worked Example

The refractive index of water is $1.33$. Find the speed of light in water. (Take $c = 3 \times 10^8\,\text{m/s}$.)

Solution

From $n = \frac{c}{v}$, we get $v = \frac{c}{n}$.
$v = \frac{3 \times 10^8}{1.33}$.
$v \approx 2.26 \times 10^8\,\text{m/s}$.

Answer: Speed of light in water $\approx 2.26 \times 10^8\,\text{m/s}$.

Worked Example

An object is placed $30\,\text{cm}$ in front of a convex lens of focal length $20\,\text{cm}$. Find the position of the image.

Solution

Sign convention: $u = -30\,\text{cm}$, $f = +20\,\text{cm}$ (convex).
Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, so $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-30}$.
$\frac{1}{v} = \frac{3 - 2}{60} = \frac{1}{60}$.
$v = +60\,\text{cm}$.

Answer: $v = +60\,\text{cm}$ — a real, inverted image is formed $60\,\text{cm}$ on the other side of the lens.

Worked Example

For the lens in Example 3 ($u = -30\,\text{cm}$, $v = +60\,\text{cm}$), find the magnification and describe the image.

Solution

For a lens, $m = \frac{v}{u} = \frac{60}{-30}$.
$m = -2$.
Negative $m$ means real and inverted; $|m| = 2 > 1$ means enlarged twice.

Answer: $m = -2$; the image is real, inverted and twice the size of the object.

Worked Example

Find the power of a convex lens of focal length $25\,\text{cm}$.

Solution

First convert focal length to metres: $f = 25\,\text{cm} = 0.25\,\text{m}$.
Power $P = \frac{1}{f} = \frac{1}{0.25}$.
$P = +4\,\text{D}$ (positive, as it is a convex lens).

Answer: Power $= +4\,\text{D}$.

Worked Example

A concave lens has a focal length of $50\,\text{cm}$. Calculate its power.

Solution

A concave lens has a negative focal length: $f = -50\,\text{cm} = -0.5\,\text{m}$.
$P = \frac{1}{f} = \frac{1}{-0.5}$.
$P = -2\,\text{D}$.

Answer: Power $= -2\,\text{D}$.

Key Points

Refraction is the bending of light at the boundary of two media due to a change in speed.
Snell's law: $\frac{\sin i}{\sin r} = n$; refractive index $n = \frac{c}{v}$.
Convex lens converges light (positive $f$, positive power); concave lens diverges light (negative $f$, negative power).
Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$; magnification $m = \frac{v}{u}$.
Power of a lens: $P = \frac{1}{f}$ (with $f$ in metres), measured in dioptre (D).

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.Light going from air into glass bends:

Explanation: Glass is denser than air, so light slows down and bends towards the normal.

Q2.The refractive index of a medium is given by:

Explanation: Refractive index $n = \frac{c}{v}$, the ratio of the speed of light in vacuum to that in the medium.

Q3.The lens formula is:

Explanation: For lenses the formula is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (note the minus sign).

Q4.The power of a convex lens of focal length $50\,\text{cm}$ is:

Explanation: $f = 0.5\,\text{m}$, so $P = \frac{1}{0.5} = +2\,\text{D}$ for a convex lens.

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