Chapter Notes

Introduction

Light is a form of electromagnetic radiation that our eyes can detect. The visible spectrum of light has wavelengths ranging from about 400 nm to 750 nm. We perceive and understand the world primarily through light and our sense of vision.

From everyday experience, we know two key things about light:

1. It travels incredibly fast.
2. It travels in a straight line.

The speed of light in a vacuum is the fastest speed possible in nature, denoted by the symbol c. Its precise value is $c=2.99792458 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$. For most calculations, we can approximate this to $c=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$.

While light is an electromagnetic wave, we can treat it as travelling in a straight line because its wavelength is extremely small compared to the size of everyday objects. This straight-line path of light is called a ray of light. A collection of such rays is called a beam of light. This "ray picture" of light is the basis of ray optics, which we use to study phenomena like reflection, refraction, and dispersion.

Reflection of Light by Spherical Mirrors

Reflection is the bouncing back of light from a surface. It is governed by two fundamental laws:

1. The angle of incidence (angle between the incident ray and the normal) is equal to the angle of reflection (angle between the reflected ray and the normal).
2. The incident ray, the reflected ray, and the normal to the surface at the point of incidence all lie in the same plane.

These laws apply to any reflecting surface, whether it's flat or curved. For spherical mirrors, the normal at any point is a line drawn from that point to the centre of curvature (the centre of the sphere from which the mirror is a part).

Key terms for spherical mirrors:

• Pole (P): The geometric centre of the spherical mirror.
• Centre of Curvature (C): The centre of the sphere of which the mirror surface is a part.
• Principal Axis: The straight line passing through the pole and the centre of curvature.

Sign convention

To use formulas for mirrors and lenses consistently, we use the Cartesian sign convention:

• Origin: All distances are measured from the pole of the mirror (or the optical centre of a lens).
• Positive Direction: The direction of the incident light is considered positive. Distances measured in this direction are positive.
• Negative Direction: The direction opposite to the incident light is considered negative. Distances measured in this direction are negative.
• Positive Height: Heights measured upwards and perpendicular to the principal axis are positive.
• Negative Height: Heights measured downwards and perpendicular to the principal axis are negative.

Focal length of spherical mirrors

When a beam of light parallel to the principal axis strikes a spherical mirror, the reflected rays behave differently for concave and convex mirrors. We assume the rays are paraxial, meaning they are close to the principal axis and make small angles with it.

• Concave Mirror: Parallel rays converge at a single point on the principal axis. This point is called the principal focus (F).
• Convex Mirror: Parallel rays appear to diverge from a single point behind the mirror on the principal axis. This point is also called the principal focus (F).

The distance between the pole (P) and the principal focus (F) is the focal length (f). For spherical mirrors, the focal length is half the radius of curvature (R).

Focal Length Formula $f = \frac{R}{2}$

The mirror equation

An image is formed where rays of light from an object meet (or appear to meet) after reflection or refraction.

• Real Image: Formed when light rays actually converge and meet at a point. It can be projected onto a screen.
• Virtual Image: Formed when light rays appear to diverge from a point but do not actually meet there. It cannot be projected onto a screen.

To find the relationship between the object distance ($u$), image distance ($v$), and focal length ($f$), we use the mirror equation.

Consider a concave mirror forming a real image A'B' of an object AB. From similar triangles in the ray diagram, we can derive the relationship.

By comparing similar triangles $\triangle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{P}$ and $\triangle \mathrm{ABP}$, and triangles $\triangle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{F}$ and $\triangle \mathrm{MPF}$ (where MP is a perpendicular from the point of incidence M to the principal axis), we arrive at the mirror equation.

The Mirror Equation $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

This equation is valid for both concave and convex mirrors, for both real and virtual images, as long as the sign convention is applied correctly.

Linear Magnification (m) is the ratio of the height of the image ($h'$) to the height of the object ($h$).

Magnification Formula $m=\frac{h^{\prime}}{h}=-\frac{v}{u}$

• If $m$ is negative, the image is inverted (real).
• If $m$ is positive, the image is erect (virtual).
• If $|m| > 1$, the image is magnified.
• If $|m| < 1$, the image is diminished.

Solution

You might think that only half the image will be formed, but that's not correct. The laws of reflection apply to every point on the mirror's surface. Rays from the object will still strike the upper half of the mirror and converge to form a complete image of the entire object.

However, since the reflecting area has been cut in half, fewer light rays will form the image. As a result, the intensity (or brightness) of the image will be reduced by half. The image of the whole object will be formed, but it will be dimmer.

Solution

The part of the phone that is perpendicular to the principal axis (like the width) will form an image of the same size on the same plane. However, the parts of the phone at different distances from the mirror along the principal axis will be magnified differently.

As seen in the ray diagram, the end of the phone closer to the mirror is magnified more than the end farther away. This is because magnification ($m = -v/u$) depends on the object distance ($u$). Since different parts of the phone have different values of $u$, their magnification will not be uniform. This causes the image to appear distorted or stretched.

Yes, the distortion will depend on the location. If the phone is placed very far from the mirror, the difference in magnification between its ends will be less significant. However, as the phone is brought closer to the mirror, especially near the focal point, the distortion becomes much more pronounced.

Given

• Radius of curvature, $R = -15 \text{ cm}$ (concave mirror)
• Focal length, $f = R/2 = -15/2 = -7.5 \text{ cm}$

To Find

Position ($v$), nature, and magnification ($m$) for: (i) Object distance, $u = -10 \text{ cm}$ (ii) Object distance, $u = -5 \text{ cm}$

Formula

Mirror Equation: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

Magnification: $m = -\frac{v}{u}$

Solution

(i) For $u = -10 \text{ cm}$

Using the mirror equation: $\frac{1}{v} + \frac{1}{-10} = \frac{1}{-7.5}$ $\frac{1}{v} = \frac{1}{10} - \frac{1}{7.5} = \frac{7.5 - 10}{75} = \frac{-2.5}{75}$ $v = -\frac{75}{2.5} = -30 \text{ cm}$

The image is formed 30 cm from the mirror on the same side as the object.

Now, calculate the magnification: $m = -\frac{v}{u} = -\frac{(-30)}{(-10)} = -3$

Answer for part (i)

• Position: 30 cm in front of the mirror.
• Nature: Real (negative $v$) and inverted (negative $m$).
• Magnification: Magnified by a factor of 3.

(ii) For $u = -5 \text{ cm}$

Using the mirror equation: $\frac{1}{v} + \frac{1}{-5} = \frac{1}{-7.5}$ $\frac{1}{v} = \frac{1}{5} - \frac{1}{7.5} = \frac{7.5 - 5}{37.5} = \frac{2.5}{37.5}$ $v = \frac{37.5}{2.5} = +15 \text{ cm}$

The image is formed 15 cm behind the mirror.

Now, calculate the magnification: $m = -\frac{v}{u} = -\frac{(+15)}{(-5)} = +3$

Answer for part (ii)

• Position: 15 cm behind the mirror.
• Nature: Virtual (positive $v$) and erect (positive $m$).
• Magnification: Magnified by a factor of 3.

Given

• Radius of curvature, $R = +2 \text{ m}$ (side view mirror is convex)
• Focal length, $f = R/2 = +1 \text{ m}$
• Speed of jogger, $v_{\text{jogger}} = 5 \text{ m s}^{-1}$

To Find

The speed of the image when the jogger is at various distances.

Formula

$v = \frac{fu}{u-f}$

Solution

The speed of the image is the change in its position per second. We can find the image position at the start and after 1 second.

(a) Jogger is between 39 m and 34 m away

• At $u_1 = -39 \text{ m}$: $v_1 = \frac{(1)(-39)}{-39-1} = \frac{-39}{-40} = \frac{39}{40} \text{ m}$
• After 1 second, the jogger is at $u_2 = -39 + 5 = -34 \text{ m}$: $v_2 = \frac{(1)(-34)}{-34-1} = \frac{-34}{-35} = \frac{34}{35} \text{ m}$
• Shift in image position in 1 s: $\Delta v = v_1 - v_2 = \frac{39}{40} - \frac{34}{35} = \frac{1365 - 1360}{1400} = \frac{5}{1400} = \frac{1}{280} \text{ m}$
• Image speed $\approx \frac{1}{280} \text{ m s}^{-1}$

(b) Jogger is between 29 m and 24 m away

• A similar calculation for $u = -29 \text{ m}$ gives an image speed of $\frac{1}{150} \text{ m s}^{-1}$.

(c) Jogger is between 19 m and 14 m away

• A similar calculation for $u = -19 \text{ m}$ gives an image speed of $\frac{1}{60} \text{ m s}^{-1}$.

(d) Jogger is between 9 m and 4 m away

• A similar calculation for $u = -9 \text{ m}$ gives an image speed of $\frac{1}{10} \text{ m s}^{-1}$.

Final Answer The speed of the image is not constant. It appears to move faster as the jogger gets closer to the mirror.

• At 39 m: $\frac{1}{280} \text{ m s}^{-1}$
• At 29 m: $\frac{1}{150} \text{ m s}^{-1}$
• At 19 m: $\frac{1}{60} \text{ m s}^{-1}$
• At 9 m: $\frac{1}{10} \text{ m s}^{-1}$

Refraction

When light passes from one transparent medium to another, it changes direction at the interface. This bending of light is called refraction. Part of the light is also reflected back into the first medium.

Refraction is governed by two laws, experimentally determined by Snell:

1. The incident ray, the refracted ray, and the normal to the interface at the point of incidence all lie in the same plane.
2. The ratio of the sine of the angle of incidence ($i$) to the sine of the angle of refraction ($r$) is a constant for a given pair of media. This is known as Snell's Law.

Snell's Law $\frac{\sin i}{\sin r}=n_{21}$ Here, $n_{21}$ is a constant called the refractive index of the second medium with respect to the first medium.

• If $n_{21} > 1$, then $r < i$. The light ray bends towards the normal. The second medium is called optically denser.
• If $n_{21} < 1$, then $r > i$. The light ray bends away from the normal. The second medium is called optically rarer.

The refractive index of medium 1 with respect to medium 2 ($n_{12}$) is the reciprocal of the refractive index of medium 2 with respect to medium 1 ($n_{21}$). $n_{12}=\frac{1}{n_{21}}$

Total Internal Reflection

When light travels from an optically denser medium to an optically rarer medium, it bends away from the normal. As the angle of incidence ($i$) increases, the angle of refraction ($r$) also increases.

At a specific angle of incidence, called the critical angle ($i_c$), the angle of refraction becomes $90^{\circ}$. At this point, the refracted ray grazes the surface of the interface.

If the angle of incidence is increased beyond the critical angle ($i > i_c$), refraction is no longer possible. Instead, all the light is reflected back into the denser medium. This phenomenon is called Total Internal Reflection (TIR). In TIR, no light is transmitted to the rarer medium, making it a perfect reflection.

From Snell's law, when $i = i_c$, $r = 90^{\circ}$. $\frac{\sin i_c}{\sin 90^{\circ}} = n_{21}$ Since $\sin 90^{\circ} = 1$, the formula for the critical angle is: $\sin i_{c}=n_{21}$ where $n_{21}$ is the refractive index of the rarer medium with respect to the denser medium.

Total internal reflection in nature and its technological applications

(i) Prisms: Prisms can be designed to use TIR to bend light by $90^{\circ}$ or $180^{\circ}$, or to invert an image without changing its size. For this to work, the critical angle of the prism's material must be less than $45^{\circ}$, which is true for common types of glass.

(ii) Optical Fibres: These are thin fibres of high-quality glass or quartz used to transmit light signals over long distances. An optical fibre consists of two parts:

• Core: The inner part, with a higher refractive index.
• Cladding: The outer layer, with a lower refractive index.

When light enters the core at a suitable angle, it strikes the core-cladding interface at an angle greater than the critical angle. It then undergoes repeated total internal reflections, travelling down the length of the fibre with almost no loss of intensity, even if the fibre is bent. This makes optical fibres essential for telecommunications and medical instruments like endoscopes.

Refraction at Spherical Surfaces and by Lenses

We can apply the laws of refraction to curved surfaces, like those found in lenses. A thin lens is an optical medium with two surfaces, at least one of which is spherical.

Refraction at a spherical surface

Consider a spherical surface with radius of curvature $R$ separating two media with refractive indices $n_1$ and $n_2$. For an object placed at distance $u$ forming an image at distance $v$, the relationship is given by:

Formula for Refraction at a Spherical Surface $\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}$ This formula is valid for any spherical surface, provided the Cartesian sign convention is used.

Given

• Object distance, $u = -100 \text{ cm}$
• Radius of curvature, $R = +20 \text{ cm}$
• Refractive index of the first medium (air), $n_1 = 1$
• Refractive index of the second medium (glass), $n_2 = 1.5$

To Find

The image position, $v$.

Formula

$\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}$

Solution

Substitute the given values into the formula: $\frac{1.5}{v}-\frac{1}{-100}=\frac{1.5-1}{20}$ $\frac{1.5}{v}+\frac{1}{100}=\frac{0.5}{20}$ $\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100} = \frac{5-2}{200} = \frac{3}{200}$ $1.5 \times 200 = 3v$ $300 = 3v$ $v = +100 \text{ cm}$

Final Answer The image is formed at a distance of 100 cm from the glass surface, inside the glass (in the direction of the incident light).

Refraction by a lens

A lens has two surfaces. We can derive a formula for the focal length of a lens by applying the spherical surface formula twice, once for each surface. The image formed by the first surface acts as the object for the second surface.

This leads to the lens maker's formula, which relates the focal length ($f$) of a lens to the refractive index of its material ($n_{21}$, relative to the surrounding medium) and the radii of curvature of its two surfaces ($R_1$ and $R_2$).

Lens Maker's Formula $\frac{1}{f}=\left(n_{21}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ This formula is used to design lenses with a desired focal length.

From the lens maker's formula, we can also derive the thin lens formula, which is analogous to the mirror equation.

Thin Lens Formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ This formula is valid for both convex (converging) and concave (diverging) lenses, and for both real and virtual images.

Magnification (m) by a Lens is the ratio of image height ($h'$) to object height ($h$). $\mathrm{m}=\frac{h^{\prime}}{h}=\frac{v}{u}$

• For an erect (virtual) image, $m$ is positive.
• For an inverted (real) image, $m$ is negative.

Solution

For the lens to "disappear," it must not bend light. This means that light passing from the liquid into the lens should not refract. Refraction only happens when light enters a medium with a different refractive index.

Therefore, the refractive index of the liquid must be the same as the refractive index of the glass lens. From the lens maker's formula, if the refractive index of the liquid ($n_1$) is equal to the refractive index of the lens ($n_2$), then $n_{21} = n_2/n_1 = 1$. $\frac{1}{f} = (1-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0$ This means $f \rightarrow \infty$, and the lens behaves like a flat sheet of glass, causing no convergence or divergence.

The refractive index of the liquid must be 1.47.

Water has a refractive index of about 1.33, so the liquid could not be water. It could be a liquid like glycerine, which has a refractive index close to 1.47.

Power of a lens

The power of a lens (P) is a measure of its ability to converge or diverge light. A lens with a shorter focal length bends light more and is considered more powerful. Power is defined as the reciprocal of the focal length (in metres).

Power of a Lens Formula $P=\frac{1}{f}$ The SI unit for power is the dioptre (D), where $1 \text{ D} = 1 \text{ m}^{-1}$.

• Converging (convex) lenses have positive power.
• Diverging (concave) lenses have negative power.

Given

• (i) $f = 0.5 \text{ m}$
• (ii) $R_1 = +10 \text{ cm}$, $R_2 = -15 \text{ cm}$, $f = +12 \text{ cm}$
• (iii) $f_{\text{air}} = +20 \text{ cm}$, $n_{\text{water}} = 1.33$, $n_{\text{glass}} = 1.5$

To Find

• (i) Power, $P$
• (ii) Refractive index of glass, $n$
• (iii) Focal length in water, $f_{\text{water}}$

Formula

$P = \frac{1}{f}$ $\frac{1}{f}=(n-1) \left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

Solution

(i) Calculate the power of the lens $P = \frac{1}{0.5 \text{ m}} = +2 \text{ D}$ Answer for part (i) The power is +2 dioptre.

(ii) Calculate the refractive index of glass Using the lens maker's formula: $\frac{1}{12} = (n-1) \left(\frac{1}{10} - \frac{1}{-15}\right)$ $\frac{1}{12} = (n-1) \left(\frac{1}{10} + \frac{1}{15}\right) = (n-1) \left(\frac{3+2}{30}\right)$ $\frac{1}{12} = (n-1) \left(\frac{5}{30}\right) = (n-1) \left(\frac{1}{6}\right)$ $n-1 = \frac{6}{12} = 0.5$ $n = 1.5$ Answer for part (ii) The refractive index of the glass is 1.5.

(iii) Calculate the focal length in water First, for the lens in air ($n_1 = 1, n_2 = 1.5$): $\frac{1}{f_{\text{air}}} = \frac{1}{20} = (1.5-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad \text{(Eq. 1)}$ Now, for the lens in water ($n_1 = 1.33, n_2 = 1.5$): $\frac{1}{f_{\text{water}}} = \left(\frac{1.5}{1.33}-1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{1.5-1.33}{1.33}\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $\frac{1}{f_{\text{water}}} = \frac{0.17}{1.33} \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \quad \text{(Eq. 2)}$ Divide Eq. 2 by Eq. 1: $\frac{1/f_{\text{water}}}{1/20} = \frac{0.17/1.33}{0.5}$ $\frac{20}{f_{\text{water}}} = \frac{0.17}{1.33 \times 0.5} = \frac{0.17}{0.665}$ $f_{\text{water}} = \frac{20 \times 0.665}{0.17} \approx 78.2 \text{ cm}$ Answer for part (iii) The focal length in water is +78.2 cm.

Combination of thin lenses in contact

If several thin lenses of focal lengths $f_1, f_2, f_3, ...$ are placed in contact, the combination acts like a single lens with an effective focal length (f).

Effective Focal Length Formula $\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}+\ldots$

In terms of power, the total power of the combination is the algebraic sum of the individual powers.

Effective Power Formula $P=P_{1}+P_{2}+P_{3}+\ldots$

The total magnification of the combination is the product of the magnifications of the individual lenses. $m=m_{1} m_{2} m_{3} \ldots$

Given

• Lens 1 (convex): $f_1 = +10 \text{ cm}$
• Lens 2 (concave): $f_2 = -10 \text{ cm}$
• Lens 3 (convex): $f_3 = +30 \text{ cm}$
• Initial object distance for Lens 1: $u_1 = -30 \text{ cm}$
• Distance between Lens 1 and 2 = 5 cm
• Distance between Lens 2 and 3 = 10 cm

To Find

The final image position, $v_3$.

Formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Solution

Step 1: Image formed by the first lens (L1) $\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10}$ $\frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{3-1}{30} = \frac{2}{30}$ $v_1 = +15 \text{ cm}$ The first image ($I_1$) is formed 15 cm to the right of L1.

Step 2: Image formed by the second lens (L2) The image $I_1$ acts as the object for L2. The distance of $I_1$ from L2 is $15 \text{ cm} - 5 \text{ cm} = 10 \text{ cm}$ to the right. Since it's on the right side, it's a virtual object for L2. So, $u_2 = +10 \text{ cm}$. $\frac{1}{v_2} - \frac{1}{+10} = \frac{1}{-10}$ $\frac{1}{v_2} = \frac{1}{10} - \frac{1}{10} = 0$ $v_2 = \infty$ The second image ($I_2$) is formed at infinity.

Step 3: Image formed by the third lens (L3) The image $I_2$ at infinity acts as the object for L3. When the object is at infinity, the image is formed at the focus. So, $u_3 = \infty$. $\frac{1}{v_3} - \frac{1}{\infty} = \frac{1}{30}$ $\frac{1}{v_3} = \frac{1}{30}$ $v_3 = +30 \text{ cm}$

Final Answer The final image is formed 30 cm to the right of the third lens.

Refraction through a Prism

A prism is a transparent optical element with flat, polished surfaces that refract light. When light passes through a triangular prism, it is deviated from its original path. The angle between the incident ray and the emergent ray is called the angle of deviation ($\delta$).

For a prism with angle $A$, the following relations hold: $r_{1}+r_{2}=A$ $\delta=i+e-A$ where $i$ is the angle of incidence, $e$ is the angle of emergence, and $r_1, r_2$ are the angles of refraction inside the prism.

The angle of deviation depends on the angle of incidence. As the angle of incidence increases, the deviation first decreases, reaches a minimum value, and then increases. This minimum value is called the angle of minimum deviation ($D_m$).

At minimum deviation, the light ray passes symmetrically through the prism, so:

• $i=e$
• $r_1 = r_2 = r$

This simplifies the equations: $2r = A \implies r = \frac{A}{2}$ $D_m = 2i - A \implies i = \frac{A+D_m}{2}$

Using Snell's law, we can find the refractive index ($n_{21}$) of the prism's material: Prism Formula $n_{21}=\frac{\sin i}{\sin r}=\frac{\sin \left(\frac{A+D_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

For a thin prism with a small angle $A$, the deviation is also small, and the formula simplifies to: $D_{m} \approx \left(n_{21}-1\right) A$

Optical Instruments

The microscope

Simple Microscope A simple microscope is just a converging (convex) lens of short focal length. It works by creating a magnified, virtual, and erect image of a small object.

• Image at Near Point (D ≈ 25 cm): To get maximum magnification, the object is placed such that the virtual image is formed at the near point of the eye (the closest distance for comfortable viewing). The magnification is: $m = 1+\frac{D}{f}$

• Image at Infinity: For more comfortable, relaxed viewing, the object is placed at the focal point ($u=f$), so the image is formed at infinity. The angular magnification is: $m = \frac{D}{f}$

Compound Microscope For much higher magnifications, a compound microscope is used. It consists of two converging lenses:

1. Objective Lens: A lens with a very short focal length ($f_o$), placed near the object. It forms a real, inverted, and magnified image.
2. Eyepiece: A lens with a short focal length ($f_e$), which acts like a simple microscope to view the image formed by the objective. It produces a final, highly magnified virtual image.

The total magnification ($m$) is the product of the magnification by the objective ($m_o$) and the eyepiece ($m_e$). For the final image formed at infinity: $m_{o} \approx \frac{L}{f_{o}} \quad \text{and} \quad m_{e} = \frac{D}{f_{e}}$ where L is the tube length, the distance between the focal points of the two lenses.

Magnification of Compound Microscope $m = m_o m_e = \frac{L}{f_{o}} \frac{D}{f_{e}}$ To achieve high magnification, both $f_o$ and $f_e$ must be very small.

Telescope

A telescope is used to view distant objects by providing angular magnification.

Refracting Telescope It consists of two converging lenses:

1. Objective Lens: Has a large focal length ($f_o$) and a large aperture (to gather more light). It forms a real, inverted image of the distant object at its focal point.
2. Eyepiece: Has a short focal length ($f_e$). It magnifies the image formed by the objective.

The magnifying power (m) of a telescope is the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the eye. Magnifying Power of Telescope $m = \frac{f_{o}}{f_{e}}$ The length of the telescope tube is $f_o + f_e$.

Reflecting Telescope Modern large telescopes use a concave mirror as the objective instead of a lens. This has several advantages:

• No Chromatic Aberration: Mirrors do not disperse light, so there are no coloured fringes in the image.
• Easier to Support: A large mirror can be supported over its entire back surface, unlike a heavy lens which can only be supported at its edges.
• Lighter and Cheaper: For the same optical quality, a mirror is lighter and less expensive to manufacture than a very large lens.

A common design is the Cassegrain telescope, which uses a large concave primary mirror and a smaller convex secondary mirror. The secondary mirror reflects light back through a hole in the primary mirror to the eyepiece. This design allows for a long focal length in a compact tube.