Lenses

(c) dioptre

The power of a lens is the reciprocal of its focal length in metres, so its unit is the dioptre (D), where $1\ \text{D} = 1\ \text{m}^{-1}$.

For a lens in a medium, the lens-maker's formula is

$\frac1f=\left(\frac{n_g}{n_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right)$

In air ($n_m=1$): $\dfrac1{24}=(1.5-1)K$.

In water ($n_m=1.33$): $\dfrac1{f_w}=\left(\dfrac{1.5}{1.33}-1\right)K$.

Dividing, $\frac{f_w}{24}=\frac{1.5-1}{\dfrac{1.5}{1.33}-1}=\frac{0.5}{0.1278}=3.91$ $f_w=24\times3.91\approx94\ \text{cm}$

The focal length of the lens in water is about $94\ \text{cm}$.

Power $P=+4$ D, so the lens is converging with focal length

$f=\frac1P=\frac14\ \text{m}=25\ \text{cm}$

Object distance $u=30$ cm (real object, so $u>f$ but $u<2f$). Using $\dfrac1v=\dfrac1f-\dfrac1u$ (real-is-positive convention):

$\frac1v=\frac1{25}-\frac1{30}=\frac{6-5}{150}\;\Rightarrow\; v=+150\ \text{cm}$

Magnification $m=-\dfrac vu=-\dfrac{150}{30}=-5$.

The image is formed $150$ cm from the lens on the side opposite the object. It is real, inverted and magnified (five times the size of the object).

Chromatic aberration: the defect of a lens by which it focuses different colours at different points, because its refractive index (hence focal length) differs for each colour. The image formed in white light is blurred and has coloured fringes. The axial chromatic aberration is $f_r-f_v$.

Proof: For a thin lens, $\dfrac1f=(n-1)K$, where $K=\dfrac1{R_1}-\dfrac1{R_2}$ is the same for all colours.

$f_v=\frac{1}{(n_v-1)K},\qquad f_r=\frac1{(n_r-1)K}$

$f_r-f_v=\frac{(n_v-1)-(n_r-1)}{(n_v-1)(n_r-1)K}=\frac{n_v-n_r}{(n_v-1)(n_r-1)K}$

Taking $(n_v-1)(n_r-1)\approx(n_y-1)^2$, where $n_y$ is the index of the mean (yellow) light,

$f_r-f_v=\frac{n_v-n_r}{n_y-1}\cdot\frac{1}{(n_y-1)K}$

The dispersive power is $\omega=\dfrac{n_v-n_r}{n_y-1}$, and $\dfrac1{(n_y-1)K}=f$, the focal length for the mean light. Therefore

$f_r-f_v=\omega f$

so the chromatic aberration equals the product of the dispersive power and the focal length for the mean light.

(a) parallel

Rays from a point at the principal focus emerge from the lens parallel to the principal axis (by reversibility of light), so the image is formed at infinity.

The lens will behave like a plain piece of glass in the liquid and will have no effect on the light.

The focal length is given by $\frac{1}{f}=\left(\frac{\mu_l}{\mu_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ where $\mu_l$ is the index of the lens and $\mu_m$ that of the surrounding medium. If $\mu_l=\mu_m$, the bracket is zero, so $1/f=0$ and $f\to\infty$. The power is zero, rays pass undeviated (no refraction at the surfaces) and the lens becomes practically invisible in the liquid.

(a) Focal length $f=\dfrac{1}{P}=\dfrac{1}{5}=0.20\ \text{m}=20\ \text{cm}$.

A real image three times the object means $v=3u$ (real-is-positive convention). Using the lens formula, $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\;\Rightarrow\;\frac{1}{u}+\frac{1}{3u}=\frac{4}{3u}=5$ $u=\frac{4}{15}\ \text{m}\approx 26.7\ \text{cm},\qquad v=3u=0.80\ \text{m}=80\ \text{cm}$

Check: $\frac{1}{0.267}+\frac{1}{0.80}=3.75+1.25=5\ \text{D}$.

(b) The power will decrease. In water the relative refractive index $\mu_g/\mu_w$ (about $1.5/1.33$) is smaller than $\mu_g/\mu_a=1.5$, so the bracket $\left(\frac{\mu_g}{\mu_m}-1\right)$ becomes smaller. The focal length increases and $P=1/f$ falls.

(a) 120 cm

$\frac{f_w}{f_a}=\frac{\mu_g-1}{\mu_g/\mu_w-1}=\frac{0.5}{1.5/1.33-1}=\frac{0.5}{0.128}\approx3.9$ $f_w\approx3.9\times30\approx117\ \text{cm}\approx120\ \text{cm}$

(a) A virtual object is a point towards which the rays are already converging when they reach the lens (for instance after another lens). The convex lens bends these converging rays even more strongly towards its axis, so they meet at a point on the far side. Rays actually pass through that point, so the image is real (and lies closer to the lens than the virtual object point).

(b)(i) The image is on the same side as the object, so it is virtual: $u=+16.0$ cm, $v=-12.0$ cm. $\frac{1}{f}=\frac1u+\frac1v=\frac1{16.0}-\frac1{12.0}=\frac{3-4}{48}=-\frac1{48}$ $f=-48\ \text{cm}$ Negative focal length: the lens is diverging (concave).

(ii) Magnification: $m=-\frac{v}{u}=-\frac{-12.0}{16.0}=+0.75$ $h'=0.75\times8.50\ \text{mm}\approx6.4\ \text{mm}$ $m$ is positive, so the image is erect, and virtual and diminished.

(iii) Ray diagram: draw the diverging lens with the principal axis, the object (8.5 mm arrow) 16 cm to its left and the focal points marked 48 cm either side. Draw (1) a ray from the object tip parallel to the axis, which diverges as if coming from the focal point on the object side; (2) a ray from the tip through the optical centre, which goes straight on. Extend the diverging rays back (dotted) on the object side. They meet 12 cm from the lens, giving an erect, smaller, virtual image there.

(i) Convex lens: A convex lens is a transparent refracting medium bounded by two spherical surfaces which bulge outwards, so that it is thicker at the middle and thinner at the edges.

Daily application: It is used as a magnifying glass (simple microscope) to see small print or objects enlarged. (It is also used in spectacles for long-sightedness, in cameras and in telescopes.)

(ii) A convex lens is called a converging lens because a parallel beam of light falling on it is refracted twice (at the two surfaces) and the rays bend towards the principal axis, meeting at a single point called the principal focus. Since the lens brings (converges) the parallel rays together instead of spreading them out, it is called a converging lens.

(iii) From the figure, the ray is incident on the first surface at an angle of incidence $i = 54^\circ$ with the normal, and it refracts into the material of the lens at an angle of refraction $r = 32^\circ$.

By Snell's law,

$n = \frac{\sin i}{\sin r}$

Substituting the values,

$n = \frac{\sin 54^\circ}{\sin 32^\circ} = \frac{0.8090}{0.5299} = 1.527$

$\boxed{n \approx 1.53}$

So the refractive index of the material of the lens is about 1.53 (the lens is made of ordinary glass).

(i) Convex lens: A convex lens is a transparent refracting medium bounded by two spherical surfaces which bulge outwards (or one spherical surface bulging outwards and the other plane surface). It is thicker at the middle and thinner at the edges, and it converges a parallel beam of light to a point, so it is also called a converging lens.

Daily application: It is used as a magnifying glass (simple microscope) to see small letters or objects enlarged.

A convex lens is called a converging lens because it refracts a parallel beam of light incident on it and converges the rays to a single point called the principal focus.

From Fig. 5.1, the ray meets the first surface with angle of incidence $i=54^\circ$ and refracts into the lens at angle of refraction $r=32^\circ$. By Snell's law,

$n=\frac{\sin i}{\sin r}=\frac{\sin54^\circ}{\sin32^\circ}=\frac{0.809}{0.530}\approx1.53$

The refractive index of the lens material is about $1.53$.

A converging lens gives an erect, magnified image only when it is virtual, formed on the same side as the object, with the object inside the focal length.

Use the real-is-positive convention: $f=+20\ \text{cm}$, object distance $u$ (positive), image distance $v=-3u$ (negative, virtual), since $m=\dfrac{|v|}{u}=3$.

$\frac1f=\frac1u+\frac1v\;\Rightarrow\;\frac1{20}=\frac1u-\frac1{3u}=\frac{2}{3u}$

$u=\frac{40}{3}\approx13.3\ \text{cm}$

Image distance: $|v|=3u=40\ \text{cm}$.

Object distance $\approx13.3\ \text{cm}$ from the lens; the image is virtual, $40\ \text{cm}$ from the lens on the same side as the object.