Chapter 9: Ray Optics
1. Introduction: The Nature of Light
Electro-magnetic radiation belongs to a vast spectrum, ranging from Gamma rays to Radio waves. The visible spectrum, which our eyes can detect, lies roughly between 400 nm (Violet) and 750 nm (Red). This is what we call "Light".
In this chapter, we ignore the wave nature of light (diffraction, interference) and assume the Rectilinear Propagation of Light. This means we assume light travels in straight lines called Rays. A bundle of such rays is called a Beam.
2. Reflection by Spherical Mirrors
A spherical mirror is simply a section cut from a hollow glass sphere, with one side polished/silvered.
(i) Concave Mirror: The outer surface is painted; reflection happens from the inner curved surface. It is Converging in nature.
(ii) Convex Mirror: The inner surface is painted; reflection happens from the outer bulging surface. It is Diverging in nature.
2.1 Important Terminology (Definitions)
- Pole (P): The geometric center of the reflecting surface (aperture) of the mirror. It serves as the origin (0,0).
- Center of Curvature (C): The center of the hollow sphere of which the mirror is a part.
- Radius of Curvature (R): The distance between Pole (P) and Center of Curvature (C).
- Principal Axis: The straight line passing through P and C.
- Principal Focus (F): A point on the principal axis where rays parallel to the axis meet (converge) or appear to diverge from after reflection.
- Focal Length (f): The distance between Pole (P) and Focus (F).
2.2 Relation between $f$ and $R$ (Derivation)
For spherical mirrors with small apertures (paraxial approximation), the focal length is half the radius of curvature.
Proof (Concave Mirror):
Consider a ray parallel to the principal axis striking the mirror at M.
Let $C$ be the center of curvature. Then $CM$ is the normal to the surface at M.
Angle of incidence $\theta = \angle MPC$.
Angle of reflection $\theta = \angle CPM$.
Since the ray is parallel to the axis, $\angle MCP = \theta$ (Alternate angles).
$\therefore \Delta CMF$ is isosceles, implying $CF = FM$.
For small aperture, $M$ is close to $P$, so $FM \approx FP$.
Thus, $CF = FP$. Since $CP = R$ and $FP = f$:
$$R = CF + FP = f + f = 2f$$
$$\Rightarrow f = \frac{R}{2}$$
3. The New Cartesian Sign Convention
To derive formulas and solve numericals, we strictly follow a coordinate geometry system where the Pole (P) is the Origin (0,0).
| Measurement | Sign Rule |
|---|---|
| Object Distance ($u$) | Always Negative (-) (Object is placed to the left) |
| Image Distance ($v$) | (-) for Real Image (Left side) (+) for Virtual Image (Right side) |
| Focal Length ($f$) | (-) for Concave Mirror (+) for Convex Mirror |
| Height ($h$) | (+) Upwards (-) Downwards (Inverted Image) |
4. The Mirror Equation & Magnification
The Mirror Formula
(Valid for both concave and convex mirrors)
4.1 Linear Magnification ($m$)
Magnification gives the extent to which the image is magnified or diminished relative to the object size.
$$m = \frac{\text{Height of Image }(h')}{\text{Height of Object }(h)}$$
In terms of object and image distances, it is derived as:
Interpreting '$m$':
• If $m$ is negative: Image is Real & Inverted.
• If $m$ is positive: Image is Virtual & Erect.
• If $|m| > 1$: Magnified.
• If $|m| < 1$: Diminished.
5. Solved Examples (Textbook Style)
Example 9.1: Suppose you are sitting in a parked car. You see a jogger approaching towards you in the side view mirror of $R = 2$ m. If the jogger is running at a speed of 5 m/s, how fast does the image of the jogger appear to move when the jogger is 39 m away?
Solution:
Given: Convex Mirror (Side view). $R = +2$ m, so $f = R/2 = +1$ m.
Object distance $u = -39$ m.
Step 1: Find Image position ($v$)
Using $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{1} - \frac{1}{-39} = 1 + \frac{1}{39} = \frac{40}{39}$
$v = \frac{39}{40}$ m.
Step 2: Differentiate to find speed
Differentiating mirror formula w.r.t time ($t$):
$-\frac{1}{v^2}\frac{dv}{dt} - \frac{1}{u^2}\frac{du}{dt} = 0$ (since $f$ is constant)
$\Rightarrow \text{Speed of Image } (\frac{dv}{dt}) = -(\frac{v}{u})^2 \times \text{Speed of Object } (\frac{du}{dt})$
$\frac{dv}{dt} = -(\frac{39/40}{-39})^2 \times 5 = -(\frac{1}{40})^2 \times 5 = -\frac{1}{1600} \times 5$
$\frac{dv}{dt} = -\frac{1}{320}$ m/s.
Answer: The image appears to move at $1/320$ m/s.
Example 9.2: An object is placed at 10 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image.
Solution:
Given: Concave Mirror. $R = -15$ cm.
Focal length $f = R/2 = -7.5$ cm.
Object distance $u = -10$ cm.
Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
$\frac{1}{v} = \frac{1}{-7.5} - \frac{1}{-10} = -\frac{10}{75} + \frac{1}{10} = -\frac{2}{15} + \frac{1}{10}$
LCM of 15 and 10 is 30.
$\frac{1}{v} = \frac{-4 + 3}{30} = -\frac{1}{30}$
$\Rightarrow v = -30$ cm.
Nature: Since $v$ is negative, the image is formed on the same side as the object (Real Image).
Magnification: $m = -v/u = -(-30)/(-10) = -3$.
The image is Real, Inverted, and Magnified 3 times.
Chapter 9: Ray Optics
1. Refraction of Light
Definition: The phenomenon of change in the path of light as it passes obliquely from one transparent medium to another is called refraction.
Cause of Refraction: The speed of light is different in different media.
• In Vacuum: $c = 3 \times 10^8$ m/s.
• In Medium: $v < c$.
When light enters a denser medium, it slows down and bends towards the normal. When it enters a rarer medium, it speeds up and bends away from the normal.
1.1 Laws of Refraction
- First Law: The incident ray, the refracted ray, and the normal to the interface at the point of incidence, all lie in the same plane.
- Second Law (Snell's Law): The ratio of the sine of the angle of incidence ($i$) to the sine of the angle of refraction ($r$) is constant for a given pair of media.
where $n_{21}$ is the refractive index of medium 2 w.r.t medium 1.
2. Real and Apparent Depth
An object placed in a denser medium (like water) appears raised when viewed from a rarer medium (like air). This is due to refraction.
[attachment_1](attachment)Refractive Index ($n$) = $\frac{\text{Real Depth}}{\text{Apparent Depth}}$
Normal Shift ($d$): The height by which the object appears raised.
$$d = \text{Real Depth} \times \left( 1 - \frac{1}{n} \right)$$
3. Total Internal Reflection (TIR)
Concept: When a light ray travels from a Denser medium to a 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 $90^\circ$ (the ray grazes the surface).
If $i > i_c$, the ray does not refract but reflects back entirely into the denser medium. This is Total Internal Reflection.
[attachment_2](attachment)Applying Snell's Law at critical angle:
$n_1 \sin i_c = n_2 \sin 90^\circ$ (where $n_1 > n_2$)
If medium 2 is air ($n_2 \approx 1$) and medium 1 is $n$:
$$\sin i_c = \frac{1}{n}$$
4. Applications of TIR
(i) Optical Fibers
Optical fibers are hair-thin strands of high-quality glass or quartz used to transmit light signals over long distances with negligible loss of energy.
- Construction: Consists of a central Core (high refractive index, $n_1$) and an outer coating called Cladding (low refractive index, $n_2$). Condition: $n_1 > n_2$.
- Working: Light entering the core at one end undergoes repeated total internal reflections and emerges at the other end.
- Uses: Telecommunications (Internet), Medical investigation (Endoscopy), Photometric sensors.
(ii) Mirage
A naturally occurring optical illusion seen in hot deserts. The air near the ground becomes hotter (rarer) than the air higher up (denser). Light from a tall object (like a tree) bends away from the normal as it moves down. Near the ground, $i > i_c$, and TIR occurs. The observer sees an inverted image, creating an illusion of water.
(iii) Totally Reflecting Prisms
Right-angled isosceles prisms ($45^\circ-90^\circ-45^\circ$) are used to deviate light by $90^\circ$ or $180^\circ$.
Reason: The critical angle for glass-air interface is $\approx 42^\circ$. In these prisms, light strikes at $45^\circ$ ($i > i_c$), ensuring TIR.
5. Solved Examples (Vol 2)
Example 9.3: Light travels from glass ($n=1.5$) to water ($n=1.33$). Calculate the critical angle for the glass-water interface.
Solution:
Condition for critical angle: $\sin i_c = \frac{n_{rarer}}{n_{denser}}$
Here, $n_{rarer} = n_w = 1.33$ and $n_{denser} = n_g = 1.5$.
$\sin i_c = \frac{1.33}{1.5} \approx 0.886$
$i_c = \sin^{-1}(0.886) \approx 62.5^\circ$.
Answer: The critical angle is $62.5^\circ$.
Chapter 9: Ray Optics
1. Refraction at a Spherical Surface
Lenses are made of spherical surfaces. Before studying lenses, we must understand how refraction happens at a single curved surface separating two media of refractive indices $n_1$ and $n_2$.
• $n_1$: Medium from which light comes (Object space).
• $n_2$: Medium into which light enters (Image space).
• $R$: Radius of curvature ($+$ or $-$ based on sign convention).
2. Refraction by Lenses
A lens is a transparent optical medium bounded by two surfaces, at least one of which is spherical.
Convex Lens: Thicker at the center, converging action.
Concave Lens: Thinner at the center, diverging action.
Objective: To find a relation between focal length ($f$), refractive index of material ($n_2$), refractive index of surrounding ($n_1$), and radii of curvature ($R_1, R_2$).
Step 1: Refraction at First Surface ($ABC$)
A ray from object $O$ in medium $n_1$ strikes the first surface ($R_1$). It refracts into medium $n_2$. If the second surface were absent, it would form a Real Image $I_1$ at distance $v_1$.
Using the spherical refraction formula:
Equation (i): $\frac{n_2}{v_1} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1}$
Step 2: Refraction at Second Surface ($ADC$)
The light now travels from $n_2$ to $n_1$. The image $I_1$ acts as a Virtual Object for this surface. The final image $I$ is formed at distance $v$.
Here: Object dist = $v_1$, Image dist = $v$.
Equation (ii): $\frac{n_1}{v} - \frac{n_2}{v_1} = \frac{n_1 - n_2}{R_2}$
Step 3: Synthesis
Adding equations (i) and (ii):
$\frac{n_2}{v_1} - \frac{n_1}{u} + \frac{n_1}{v} - \frac{n_2}{v_1} = \frac{n_2 - n_1}{R_1} + \frac{n_1 - n_2}{R_2}$
$\Rightarrow \frac{n_1}{v} - \frac{n_1}{u} = (n_2 - n_1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
Dividing both sides by $n_1$:
$\frac{1}{v} - \frac{1}{u} = (n_{21} - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$
Step 4: Final Result
If object is at infinity ($u = \infty$), image forms at focus ($v = f$). Thus, $1/v - 1/u = 1/f$.
3. Thin Lens Formula & Magnification
Just like mirrors, lenses follow a formula relating $u, v, f$.
Note: Notice the negative sign (unlike mirrors).
For Convex Lens: $f$ is Positive ($+$).
For Concave Lens: $f$ is Negative ($-$).
3.1 Power of a Lens ($P$)
Power is a measure of the degree of convergence or divergence a lens provides.
- Definition: The tangent of the angle by which it converges a light beam falling at unit distance from the optical center.
- Formula: $P = \frac{1}{f(\text{meters})}$.
- SI Unit: Diopter (D). ($1D = 1 m^{-1}$).
- Sign: Power of Convex lens is (+), Concave lens is (-).
3.2 Combination of Lenses in Contact
If two thin lenses of focal lengths $f_1$ and $f_2$ are placed in contact, they behave as a single equivalent lens.
4. Solved Examples (Vol 3)
Example 9.4: A double convex lens is to be manufactured from glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Solution:
Given: $\mu = 1.55$, $f = +20$ cm.
Since it is double convex and symmetric: $R_1 = +R$ and $R_2 = -R$.
Using Lens Maker's Formula:
$\frac{1}{f} = (\mu - 1) [\frac{1}{R_1} - \frac{1}{R_2}]$
$\frac{1}{20} = (1.55 - 1) [\frac{1}{R} - \frac{1}{-R}]$
$\frac{1}{20} = 0.55 \times [\frac{2}{R}]$
$\frac{1}{20} = \frac{1.1}{R}$
$R = 20 \times 1.1 = 22$ cm.
Answer: The required radius of curvature is 22 cm.
Chapter 9: Ray Optics
1. Refraction Through a Prism
A prism is a homogeneous, transparent medium enclosed by two plane refracting surfaces inclined at an angle called the Angle of Prism ($A$).
[attachment_0](attachment)Geometric Relations (Derivation Logic)
Consider a ray entering the prism at angle $i$, refracting at $r_1$, striking the second face at $r_2$, and emerging at angle $e$.
From the quadrilateral formed by normals: $\angle A + \angle QNR = 180^\circ$.
From the triangle formed by refracted ray: $r_1 + r_2 + \angle QNR = 180^\circ$.
Comparing both: $r_1 + r_2 = A$.
The total deviation $\delta$ is the sum of deviations at two surfaces:
$\delta = (i - r_1) + (e - r_2)$
$\delta = (i + e) - (r_1 + r_2)$
$\delta = i + e - A$
2. Minimum Deviation ($\delta_m$)
If we plot a graph between the angle of incidence ($i$) and angle of deviation ($\delta$), we get a U-shaped curve (Parabola). The deviation is minimum at only one specific value of $i$.
[attachment_1](attachment)1. The ray passes symmetrically through the prism.
2. Angle of Incidence = Angle of Emergence ($i = e$).
3. Internal angles are equal ($r_1 = r_2 = r = A/2$).
4. Refracted ray is parallel to the base of the prism.
2.1 The Prism Formula
Substituting values at minimum deviation:
$A + \delta_m = i + i \Rightarrow i = \frac{A + \delta_m}{2}$
$r = A/2$
Using Snell's Law ($n = \sin i / \sin r$):
3. Dispersion of Light
The splitting of white light into its constituent colors (VIBGYOR) is called Dispersion.
[attachment_2](attachment)Cause: The refractive index of glass is different for different wavelengths (Cauchy's Formula).
$\mu = A + \frac{B}{\lambda^2}$
Since $\lambda_{red} > \lambda_{violet}$, therefore $\mu_{red} < \mu_{violet}$.
Result: Violet bends the most, Red bends the least.
4. Natural Phenomena (Scattering)
4.1 The Rainbow
A combined effect of Dispersion, Refraction, and Internal Reflection of sunlight by spherical water droplets.
Primary Rainbow: Two refractions + One internal reflection. (Red outer, Violet inner).
Secondary Rainbow: Two refractions + Two internal reflections. (Violet outer, Red inner).
4.2 Scattering of Light
Rayleigh's Law: The intensity of scattered light ($I$) is inversely proportional to the fourth power of wavelength.
$$I \propto \frac{1}{\lambda^4}$$
- Blue Sky: Blue light has a shorter wavelength, so it scatters nearly 10 times more than red light. This scattered light enters our eyes.
- Red Sunset: At sunrise/sunset, light travels a longer distance through the atmosphere. Most blue light scatters away. Only the red light (longer $\lambda$) reaches our eyes.
5. Solved Examples (Vol 4)
Example 9.5: A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be $40^\circ$. What is the refractive index of the material of the prism? The refracting angle of the prism is $60^\circ$.
Solution:
Given: $A = 60^\circ$, $\delta_m = 40^\circ$.
Using Prism Formula:
$\mu = \frac{\sin(\frac{60 + 40}{2})}{\sin(\frac{60}{2})} = \frac{\sin 50^\circ}{\sin 30^\circ}$
$\mu = \frac{0.766}{0.5} = 1.532$
Answer: The refractive index is 1.532.
Chapter 9: Ray Optics
1. Refraction Through a Prism
A prism is a homogeneous, transparent medium enclosed by two plane refracting surfaces inclined at an angle called the Angle of Prism ($A$).
[attachment_0](attachment)Geometric Relations (Derivation Logic)
Consider a ray entering the prism at angle $i$, refracting at $r_1$, striking the second face at $r_2$, and emerging at angle $e$.
From the quadrilateral formed by normals: $\angle A + \angle QNR = 180^\circ$.
From the triangle formed by refracted ray: $r_1 + r_2 + \angle QNR = 180^\circ$.
Comparing both: $r_1 + r_2 = A$.
The total deviation $\delta$ is the sum of deviations at two surfaces:
$\delta = (i - r_1) + (e - r_2)$
$\delta = (i + e) - (r_1 + r_2)$
$\delta = i + e - A$
2. Minimum Deviation ($\delta_m$)
If we plot a graph between the angle of incidence ($i$) and angle of deviation ($\delta$), we get a U-shaped curve (Parabola). The deviation is minimum at only one specific value of $i$.
1. The ray passes symmetrically through the prism.
2. Angle of Incidence = Angle of Emergence ($i = e$).
3. Internal angles are equal ($r_1 = r_2 = r = A/2$).
4. Refracted ray is parallel to the base of the prism.
2.1 The Prism Formula
Substituting values at minimum deviation:
$A + \delta_m = i + i \Rightarrow i = \frac{A + \delta_m}{2}$
$r = A/2$
Using Snell's Law ($n = \sin i / \sin r$):
3. Dispersion of Light
The splitting of white light into its constituent colors (VIBGYOR) is called Dispersion.
Cause: The refractive index of glass is different for different wavelengths (Cauchy's Formula).
$\mu = A + \frac{B}{\lambda^2}$
Since $\lambda_{red} > \lambda_{violet}$, therefore $\mu_{red} < \mu_{violet}$.
Result: Violet bends the most, Red bends the least.
4. Natural Phenomena (Scattering)
4.1 The Rainbow
A combined effect of Dispersion, Refraction, and Internal Reflection of sunlight by spherical water droplets.
Primary Rainbow: Two refractions + One internal reflection. (Red outer, Violet inner).
Secondary Rainbow: Two refractions + Two internal reflections. (Violet outer, Red inner).
4.2 Scattering of Light
Rayleigh's Law: The intensity of scattered light ($I$) is inversely proportional to the fourth power of wavelength.
$$I \propto \frac{1}{\lambda^4}$$
- Blue Sky: Blue light has a shorter wavelength, so it scatters nearly 10 times more than red light. This scattered light enters our eyes.
- Red Sunset: At sunrise/sunset, light travels a longer distance through the atmosphere. Most blue light scatters away. Only the red light (longer $\lambda$) reaches our eyes.
5. Solved Examples (Vol 4)
Example 9.5: A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be $40^\circ$. What is the refractive index of the material of the prism? The refracting angle of the prism is $60^\circ$.
Solution:
Given: $A = 60^\circ$, $\delta_m = 40^\circ$.
Using Prism Formula:
$\mu = \frac{\sin(\frac{60 + 40}{2})}{\sin(\frac{60}{2})} = \frac{\sin 50^\circ}{\sin 30^\circ}$
$\mu = \frac{0.766}{0.5} = 1.532$
Answer: The refractive index is 1.532.
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