Class 10 Science Chapter 9 Light – Reflection and Refraction

Key Concept: Symbol Representation

d) F
[Solution Description]
In optics, the focal length is commonly represented by the letter $ f $. It is an essential parameter in determining the behavior of lenses and mirrors.

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Key Concept: Basic Concept

c) f
[Solution Description]
In optics, the letter  f  is commonly used to represent the focal length of a lens. This notation helps identify and distinguish the focal length from other optical properties.

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Key Concept: Basic Concept

c) The point where parallel incident rays meet after reflection
[Solution Description] The principal focus of a concave mirror is the point on the principal axis where light rays parallel to the principal axis converge after reflection.

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Key Concept: Combined Lenses, Magnification Calculation

a) 4
[Solution Description] First, determine the equivalent focal length $F$ of the lens system using the formula:$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
where $f_1 = 10\ \text{cm}$ and $f_2 = -20\ \text{cm}$. Substituting the values, we have:$\frac{1}{F} = \frac{1}{10} + \left(-\frac{1}{20}\right)$
Simplifying gives:$\frac{1}{F} = \frac{2}{20} – \frac{1}{20} = \frac{1}{20}$$So, $F = 20\ \text{cm}$.

Next, use the lens formula for the combined system to find the image distance $v$:$\frac{1}{v} – \frac{1}{u} = \frac{1}{F}$
Given $u = -15\ \text{cm}$, substitute the known values:$\frac{1}{v} – \frac{1}{-15} = \frac{1}{20}$
Simplify to find $v$:$\frac{1}{v} = \frac{1}{20} – \frac{1}{15} = \frac{3 – 4}{60} = -\frac{1}{60}$
Thus, $v = -60\ \text{cm}$.

Finally, calculate the magnification $m$:$m = \frac{v}{u} = \frac{-60}{-15} = 4$

Notes:

• Combined Lenses:
  • When two lenses are placed close to each other, their combined focal length is given by:
    $1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$​
    where F$F$ is the combined focal length, f1f_1and f2f_2are the focal lengths of individual lenses.
  • The combination behaves as a single lens with the focal length
    F.
• Magnification Calculation:
  • Magnification ($M$$M=h′h=−vuM = \frac{h’}{h} = \frac{-v}{u}$​
    where h is the image height, h$h$ is the object height,
  • v is the image distance, and
  • u is the object distance.
  • For combined lenses, calculate
    $v$ using the effective F
  • , then determine M.
  • Applications: Used in microscopes, telescopes, and optical instruments

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Key Concept: Principal Focus

c) Principal focus
[Solution Description] The point where parallel rays of light converge after passing through a convex lens is known as the principal focus of the lens. This is due to the lens’s ability to refract light towards this focal point.

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Key Concept: Complex Ray Diagrams, Comparative Refraction, Numerical and Analytical Difficulty

d) 30 cm to the left
[Solution Description] First, consider the effect of the convex lens. Since the object is at infinity, the image formed by the convex lens will be at its focal point (10 cm from the lens) on the opposite side. This image acts as a virtual object for the concave lens.

The distance of this image to the concave lens (u) is given by:$ u = 20 \, \text{cm} – 10 \, \text{cm} = 10 \, \text{cm} $

Using the lens formula for the concave lens:$\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$

Substitute the values ($f = -15 \, \text{cm}$, and $u = -10 \, \text{cm}$):$\frac{1}{-15} = \frac{1}{v} + \frac{1}{10}$

Solving for v:$\frac{1}{v} = \frac{1}{-15} – \frac{1}{10}$ $\frac{1}{v} = \frac{-2 + 3}{30}$ $\frac{1}{v} = \frac{1}{30}$ $v = 30 \, \text{cm}$

Thus, the final image is 30 cm on the left side of the concave lens.

Notes:

• Complex Ray Diagrams: Involves tracing multiple rays to understand image formation in lenses and mirrors. It includes both paraxial and non-paraxial rays for detailed analysis.
• Comparative Refraction: Refraction occurs when light passes through media of different densities. It is compared in terms of angles of incidence and refraction, using Snell’s Law.
• Numerical Problems: Common in refraction and lens formula applications (1/f = 1/v – 1/u). Involves calculating focal length, object/image distances, magnification.
• Analytical Difficulty: Involves solving complex problems using ray diagrams, equations of refraction, and understanding deviation, focal length, and image formation

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Key Concept: Intermediate Calculation

c) 1.67

[Solution Description]

The refractive index ($n$) of a medium can be calculated using the formula:

$n = \frac{c}{v}$

where $c = 3 \times 10^8$ m/s is the speed of light in vacuum, and $v = 1.8 \times 10^8$ m/s is the speed of light in the medium.

Substituting these values into the formula:

$n = \frac{3 \times 10^8}{1.8 \times 10^8}$

Simplifying the fraction gives:

$n = \frac{3}{1.8}$

$n = 1.67$

Therefore, the refractive index of the medium is 1.67.

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Key Concept: Mirror Usage

c) They have a wider field of view
[Solution Description] Convex mirrors are commonly used as rear-view mirrors because they provide a wider field of view than plane mirrors. This allows drivers to see more of the traffic behind them.

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Key Concept: Magnification Concept

b) Image is real and same size
[Solution Description]
Magnification m given by a mirror is $$\frac{-v}{u}$$ for a concave mirror, where v is the image distance and u is the object distance. A magnification of -1 indicates that the image’s height is equal to the object’s height but inverted. Thus, the image is real, inverted, and the same size as the object.

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Key Concept: Complex Problem Solving, Advanced Application

a) Virtual, diminished, and smaller by a factor of 0.25
[Solution Description] To solve this problem, we use the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ where  v  is the image distance,  u  is the object distance, and  f  is the focal length.

For a convex mirror, the focal length is positive, so $ f = +4 \, \text{m} $. The object distance $ u = -12 \, \text{m} $ as per the sign convention.

Plug these values into the mirror formula:

$\frac{1}{v} = \frac{1}{f} – \frac{1}{u} = \frac{1}{4} – \frac{1}{-12} = \frac{1}{4} + \frac{1}{12}$

Finding a common denominator,

$\frac{1}{v} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$

Thus,

$v = 3 \, \text{m}$

The image is virtual, erect, and diminished. Next, calculate magnification using $ m = -\frac{v}{u} $:

$m = -\frac{3}{-12} = \frac{1}{4} = 0.25$

Therefore, the image is reduced by a factor of 0.25.

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Key Concept: Conceptual Understanding, Advanced Real-World Application

c) Because it refracts light rays diversely causing virtual focus
[Solution Description]
A concave lens diverges light rays because its shape causes parallel incident rays to spread out. When these rays are extended backward, they appear to emanate from a single point. This point lies on the same side of the lens as the object itself, thus forming a virtual image.

In practical applications, such as eyeglasses for nearsightedness, concave lenses help diverge incoming light before it reaches the eye, adjusting the focal point onto the retina and helping individuals see distant objects more clearly. Since the lens only alters the perceived origin of light, it can never converge the rays to form a real image; hence, all images created are upright, smaller than the object, and virtual.

Notes:

Conceptual Understanding:

1. Chapter 9 delves into the foundational principles of light, including reflection and refraction.
2. Discusses laws of reflection and refraction, Snell’s law, and refractive index.
3. Explores ray diagrams for lenses and their use in magnification.
4. Introduces concepts of the power of a lens and optical instruments.

Advanced Real-World Applications:

1. Application of lenses in devices like microscopes and telescopes.
2. Use of mirrors in solar panels and optical imaging systems.
3. Practical use of refraction in spectacles, cameras, and projectors.
4. Concepts underpin laser technology and fiber optics in modern communication.

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Key Concept: Sign Convention Basics

b) Negative
[Solution Description] The New Cartesian Sign Convention states that the focal length for a concave mirror is negative because the principal focus lies on the side opposite to the direction of the incident light.

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Key Concept: Complex Calculation, Numerical and Analytical Difficulty

c) 60 cm, real, inverted
[Solution Description]
To find the image distance $v$, we will use the mirror formula:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f},$
where $u$ is the object distance and $f$ is the focal length. Here, $u = -30\, \text{cm}$ (since it is on the same side as the incoming light), and $f = -20\, \text{cm}$ (negative for concave mirrors).

Substituting the values into the formula:
$\frac{1}{v} = \frac{1}{-20} – \frac{1}{-30}.$

Simplifying gives:
$\frac{1}{v} = \frac{-3 + 2}{60} = \frac{-1}{60}.$

Hence,
$v = -60\, \text{cm}.$

The negative sign indicates that the image is real and inverted, located 60 cm from the mirror on the same side as the object.

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Key Concept: Ray Path

a) Along the same path
[Solution Description] When a ray passes through the center of curvature of a concave mirror, it falls on the mirror normally and thus reflects back along the same path as per the law of reflection.

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Key Concept: Image Formation

c) Beyond the center of curvature
[Solution Description] For a concave mirror, when the object is placed beyond the center of curvature (C), the image formed is real, inverted, and diminished. This is because the rays converge after reflecting from the mirror and meet at a point between the focus (F) and C on the principal axis.

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Key Concept: Real-World Refraction

b) Due to refraction
[Solution Description] The apparent bending of the pencil occurs due to the phenomenon of refraction. When light passes from water (denser medium) to air (rarer medium), it speeds up and bends away from the normal. This change in direction causes the submerged part of the pencil to appear displaced from its actual position, creating an illusion of bending.

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Key Concept: Image Size

b) Diminished
[Solution Description] The image formed by a convex mirror is always virtual, erect, and diminished in size.

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Key Concept: Complex Calculations, Optical Density

c) 125 cm
[Solution Description]
We use the Lensmaker’s formula for a plano-convex lens:$\frac{1}{f} = (n_{\text{{glass}}} – n_{\text{{medium}}}) \left(\frac{1}{R_1} – \frac{1}{R_2}\right)$

For a plano-convex lens, $R_1 = \infty$ (for the flat side), and $R_2 = -10$ cm. Substituting these values and the given refractive indices:$
$\frac{1}{f} = (1.52 – 1.44) \left(\frac{1}{\infty} + \frac{1}{10}\right)$

Simplifying:$\frac{1}{f} = 0.08 \times \frac{1}{10}$ $\frac{1}{f} = 0.008$

Therefore, $f = \frac{1}{0.008}$ $f = 125 \, \text{cm}$

Hence, the focal length of the lens is 125 cm.

Notes:

• Optical Density (OD): It refers to the measure of how much a medium can refract light. Higher optical density means light slows down more when passing through the medium.
• Formula: Optical Density (n) = Speed of Light in Vacuum (c) / Speed of Light in Medium (v)
• Refractive Index (n): This is a measure of how much light bends when entering a new medium. It is calculated using Snell’s Law:
  $$n = \frac{\sin i}{\sin r}$$, where i is the angle of incidence, and r is the angle of refraction.
• Light Behavior: In optically denser mediums, light bends towards the normal. In less dense mediums, it bends away from the normal.
• Applications: Used in designing lenses, prisms, and understanding light travel in different materials

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Key Concept: Mirror Type

c) Convex mirror
[Solution Description] Among the different types of mirrors, a convex mirror always forms a virtual and erect image regardless of the position of the object. When the object is at infinity, the rays appear to come from the focus behind the mirror, giving a virtual and erect image.

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Key Concept: Lens Formula Application

c) -6.67 cm
[Solution Description]
To solve this, we use the lens formula:
$\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$

where $u = -20$ cm, $f = -10$ cm. Substituting these into the formula gives:
$\frac{1}{v} + \frac{1}{20} = \frac{-1}{10}$
$\frac{1}{v} = \frac{-1}{10} – \frac{1}{20}$
$\frac{1}{v} = \frac{-2 – 1}{20} = \frac{-3}{20}$
Therefore, solving for $v$, we get:
$v = \frac{20}{-3} = -6.67 \text{ cm}$
The image distance is -6.67 cm. The negative sign indicates that the image is virtual and on the same side as the object.

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Key Concept: Optical Density and Mass Density, Complex Refraction Path

b) 36.93°
[Solution Description]
Start by analyzing the air to sapphire transition:

For air to sapphire: $n_{air} \sin(i_1) = n_{sapphire} \sin(r_1)$,
with $n_{air} = 1$, $i_1 = 60$, $n_{sapphire} = 1.77$.

Calculating $\sin(r_1)$:\sin(r_1) &= \frac{\sin(60°)}{1.77} \\&= \frac{0.8660}{1.77} \\&\approx 0.4893.
Thus, $r_1 \approx \arcsin(0.4893) \approx 29.35°$.

Now proceed to sapphire to kerosene transition:
Sapphire to Kerosene: $n_{sapphire} \sin(r_1) = n_{kerosene} \sin(r_2)$,
where $r_1 \approx 29.35°$, $n_{kerosene} = 1.44$.

Solving for $\sin(r_2)$:$\sin(r_2) = \frac{1.77 \times \sin(29.35°)}{1.44} \approx \frac{1.77 \times 0.4893}{1.44} \approx 0.6019.$
Therefore, $r_2 \approx \arcsin(0.6019) \approx 36.93°$.

The exit angle in kerosene is approximately 36.93°.

Notes:

• Optical Density: It refers to the ability of a medium to bend light. Higher optical density means light travels slower, and the light bends more when entering that medium.
• Mass Density: This is the mass per unit volume of a substance, expressed in kg/m³.
• Relation Between Optical and Mass Density: Optical density depends on the mass density and the refractive index of the medium.
• Refraction Path: When light passes through different mediums, it bends, creating a complex refraction path.
• Snell’s Law: Governs the refraction of light and is essential for understanding the bending of light in various media

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Key Concept: Lens Power Calculation, Complex Lens Formula Problems

b) 10.42 diopters
[Solution Description]
Using the lens formula:$\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$,

where $ u = -12 \, \text{cm} $ and $ v = 48 \, \text{cm} $.\frac{1}{f} & = \frac{1}{48} + \frac{1}{12} \\& = \frac{1 + 4}{48} = \frac{5}{48}.

Thus, $ f = \frac{48}{5} = 9.6 \, \text{cm} $.

The power $ P $ of the lens is given by $ P = \frac{100}{f (\text{in cm})} $.P & = \frac{100}{9.6} \\& = 10.42 \, \text{diopters}.

Therefore, the power of the lens is approximately 10.42 diopters.

Notes:

• Lens Power:
  • Power (P) of a lens is measured in dioptres (D).
  • Formula:
    P=100f(in cm)P = \frac{100}{f(\text{in cm})}P=f(in cm)100​, where
  • f is the focal length in meters.
  • Convex lens: Positive power; Concave lens: Negative power.
• Lens Formula:
  • 1f=1v−1u\frac{1}{f} = \frac{1}{v} – \frac{1}{u}, where f is focal length,
  • v is image distance,
  • u is object distance.
• Combination of Lenses:
  • Combined power:
    Ptotal=P1+P2+…P_{\text{total}} = P_1 + P_2 + \dots
  • Effective focal length:
    1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}
• Complex Problems:
  • Apply sign conventions carefully.
  • Use lens formula and magnification equations systematically.
  • Solve step-by-step for clarity

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Key Concept: Combined Formula Application

b) 0.667

[Solution Description] Use the mirror formula first:

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.

Substituting the given values:

$\frac{1}{v} + \frac{1}{10} = \frac{1}{-20}$.

Solve for $\frac{1}{v}$:

\[\frac{1}{v} = \frac{1}{-20} – \frac{1}{10}
= \left( \frac{-1 – 2}{20} \right)
= \frac{-3}{20}.\]

Hence,

$v = -\frac{20}{3} \approx -6.67 \, \text{cm}.$

Now find the magnification using $m = -\frac{v}{u}$:

$m = -\left(\frac{-6.67}{10}\right) = 0.667.$

Thus, the magnification is approximately 0.667.

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Key Concept: Complex Ray Path, Angle Calculation

d) 30 degrees
[Solution Description]
The initial incident angle on the concave mirror is 60 degrees with respect to the principal axis. Thus, by the law of reflection, it will reflect back at the same angle relative to the principal axis. When it strikes the plane mirror, the angle with the normal is given as 30 degrees. By the law of reflection, the angle of incidence equals the angle of reflection, so the angle of reflection from the plane mirror will also be 30 degrees.

Therefore, after reflecting off the concave mirror, the ray travels to the plane mirror where it reflects symmetrically with reference to the normal.

Hence, the answer is 30 degrees.

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Key Concept: Real vs Virtual Images

a) Can be caught on a screen

[Solution Description]

Real images are formed when light rays actually converge at a point. They can be projected onto a screen, unlike virtual images, which appear to form behind the mirror and cannot be captured on a screen.

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Key Concept: Normal line

c) It provides a reference line to measure the angle of incidence and reflection.
[Solution Description] The normal line at the point of incidence is crucial because it helps determine the angles of incidence and reflection. For any reflective surface, the law of reflection states that the angle of incidence equals the angle of reflection, both measured with respect to the normal.

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Key Concept: Basic Definitions

b) Convex mirror
[Solution Description] A spherical mirror whose reflecting surface is curved outwards, facing away from the center of the sphere, is known as a convex mirror.

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Key Concept: Real-World Application

d) Beyond $2F_1$
[Solution Description]
For the image to be smaller than the actual object when using a convex lens, the object must be beyond twice the focal length $(2F_1)$ This is based on the properties of lenses which state that:

– When an object is placed between $F_1$ and $2F_1$, the image is real, inverted, and larger.
– Placed at $2F_1$, the image is the same size.
– Beyond $2F_1$, the image is diminished (smaller).

Thus, placing the object beyond $2F_1$ results in a smaller, diminished, real, and inverted image.

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Key Concept: Real-World Application, Advanced Conceptual Understanding

c) 60 degrees
[Solution Description] The problem can be solved using Snell’s law, $n_1 \sin i = n_2 \sin r$, twice: once for entering and once for exiting the medium.
For entrance to the medium,
$n_{\text{air}} \sin(60^\circ) = 1.5 \sin(r_1)$
where $n_{\text{air}} = 1$.
$\sin(r_1) = \frac{\sin(60^\circ)}{1.5} = \frac{\sqrt{3}/2}{1.5} \approx 0.577$
Solving, we find $\approx 35.26^\circ$.
When light emerges back into air,
$1.5 \sin(r_2) = n_{\text{air}} \sin(60^\circ)$
$\sin(r_2) = \frac{\sqrt{3}/2}{1.5} \approx 0.577$
Hence, $r_2 \approx 60^\circ$. Thus, the light exits at an angle of 60 degrees.

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Key Concept: Complex Angle Calculation, Refraction Path Prediction

c) 19.47 degrees
[Solution Description]
We use Snell’s Law to calculate the angle of refraction:$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$

where $ n_1 $ is the refractive index of air (approximately 1), $ \theta_1 $ is the angle of incidence (30 degrees), $ n_2 $ is the refractive index of glass (1.5), and $ \theta_2 $ is the angle of refraction. Substituting the known values:$1 \times \sin(30^\circ) = 1.5 \times \sin(\theta_2)$

Solving for $ \sin(\theta_2) $:$\sin(\theta_2) = \frac{1 \times 0.5}{1.5} \approx 0.3333$

Using inverse sine function:$\theta_2 \approx \sin^{-1}(0.3333) \approx 19.47^\circ$

Thus, the angle of refraction is approximately 19.47 degrees.

Notes:

• Refraction of Light: The bending of light when it passes from one medium to another due to a change in speed.
• Snell’s Law: Describes refraction mathematically:
  $n_1 \sin \theta_1 = n_2 \sin \theta_2$​.
  • $n_1 \text{ and } n_2$
     are refractive indices of the two media.
  • $\theta_1 \text{ and } \theta_2$
    are angles of incidence and refraction, respectively.
• Critical Angle: The angle of incidence that results in a refraction angle of 90°.
• Total Internal Reflection: Occurs when the angle of incidence exceeds the critical angle.
• Path Prediction: By using Snell’s law and refractive indices, one can predict the direction of light after refraction

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Key Concept: Sign Convention Application

b) -6.67 cm

[Solution Description]

To find the image distance (v), we use the lens formula:

$\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$.

Here, f = -10 cm and u = -20 cm (as object distance is negative).

Plugging values into the formula:

$\frac{1}{v} – \frac{1}{-20} = \frac{1}{-10}$

Simplifying gives:

$\frac{1}{v} + \frac{1}{20} = -\frac{1}{10}$

$\frac{1}{v} = -\frac{1}{10} – \frac{1}{20} = -\frac{2}{20} – \frac{1}{20} = -\frac{3}{20}$

$v = -\frac{20}{3} \approx -6.67 \text{ cm}$

So, the image distance v is approximately -6.67 cm.

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Key Concept: Advanced Calculations, Conceptual Application

b) Image is virtual, at 48 cm from the mirror
[Solution Description]
The formula to find the focal length f of a spherical mirror is $f = \frac{R}{2}$, where $R$ is the radius of curvature. Substituting $R = 24 \text{ cm}$, we get $f = \frac{24}{2} = 12 \text{ cm}$.

Using the mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, where v is the image distance and u is the object distance:

$\frac{1}{v} = \frac{1}{12} – \frac{1}{16}
= \frac{16 – 12}{192}
= \frac{4}{192}
= \frac{1}{48}.$

Solving for v, we find $v = 48 \text{ cm}$. Since v is positive, the image is virtual and located behind the mirror.

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Key Concept: Angle of Deviation, Reflection and Refraction

c) 0 degrees
[Solution Description] When light passes from air (refractive index 1) into glass (refractive index 1.5), it slows down and bends towards the normal. According to Snell’s law:
$n_1 \sin i = n_2 \sin r$
Here, $n_1 = 1$, $i = 30^\circ$, and $n_2 = 1.5$. Therefore,
$\sin r = \frac{\sin 30}{1.5} = \frac{0.5}{1.5} \approx 0.333$
Hence, $r \approx \arcsin(0.333) \approx 19.47^\circ$.

As the light exits the glass, it refracts again:
$n_2 \sin r’ = n_1 \sin e$
Solving for $e$:
$\sin e = 1.5 \times \sin 19.47 \approx 0.500 \quad \Rightarrow \quad e \approx 30^\circ$

The angle of deviation ($$\delta$$) is given by:
$\delta = e – i = 30^\circ – 30^\circ = 0^\circ$
Thus, there is no angular deviation as the beam emerges parallel to the incident direction.

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Key Concept: Conceptual Understanding, Non-Straightforward Path

b) It is negative and equal to -0.83 D.
[Solution Description]
The power Pof a lens is given by$P = \frac{1}{f}$

For a concave lens, the focal length f is negative. Therefore,$P = \frac{1}{-1.2} = -\frac{5}{6} \approx -0.83 \text{ D}$

This indicates that the power of the lens is negative, typical of concave lenses.

Notes:

• Definition: A non-straightforward path refers to the actual distance covered by an object, which may differ from the shortest distance (displacement) between two points.
• Distance vs. Displacement:
  • Distance: Total length of the path traveled, irrespective of direction.
  • Displacement: Shortest straight line from start to endpoint with direction.
• Key Concept: For a non-straightforward path:
  • Distance > Displacement (except in straight-line motion).
• Examples:
  • Circular paths, zigzag roads, or curved trajectories.
• Applications: Understanding helps in analyzing real-world motion like vehicle movement or river flow.

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Key Concept: Real-World Application, Multiple Ray Analysis

b) 20.04 cm behind the lens
[Solution Description]
For a convex lens, we use the lens formula:
$\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$

Where $f = 20\,cm$, $u = -10000\,cm$ (since the object is far away, convert meters to centimeters), and $v$ is the image distance.

First, rearrange the formula to solve for $v$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-10000}$

Calculate:
$\frac{1}{v} = \frac{1}{20} – \frac{1}{10000} = \frac{500}{10000} – \frac{1}{10000} = \frac{499}{10000}$
$v = \frac{10000}{499} \approx 20.04\,cm$

Thus, the image forms approximately 20.04 cm behind the lens.

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Key Concept: Theoretical Analysis, Advanced Reasoning

a) The effective focal length is infinite (behaves like no lens).
[Solution Description]
For lenses in contact, the total power $ P $ is the sum of the individual powers:$
P = P_{\text{convex}} + P_{\text{concave}}$

Convert focal lengths to powers:$P_{\text{convex}} = \frac{1}{f_{\text{convex}}} = \frac{1}{0.2} = 5D$

$P_{\text{concave}} = \frac{1}{f_{\text{concave}}} = -\frac{1}{0.2} = -5D$

Calculate total power:$P = 5D – 5D = 0D$

A net power of zero means the system’s effective focal length is infinite, behaving like a flat piece of glass.

Notes:

• Reflection and Refraction: Light behaves differently when it strikes different surfaces—reflection occurs when light bounces off, while refraction happens when light passes through materials of different densities.
• Lens and Mirrors: The behavior of light through lenses (concave and convex) and mirrors (concave and convex) is essential for understanding image formation.
• Human Eye: The structure and function of the human eye, including the role of the lens, retina, and cornea in vision.
• Color Theory: The dispersion of light into colors through prisms and the formation of the rainbow.
• Optical Instruments: Microscopes, telescopes, and other instruments that manipulate light to magnify images for better visibility

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Key Concept: Lens System Analysis, Complex Lens Formula Problems

b) Effective focal length: -40 cm, Diverging
[Solution Description] When dealing with systems of separated lenses, calculate the effective focal length using:$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} – \frac{d}{f_1 f_2}$

where $ f_1 = +20 \, \text{cm}, f_2 = -10 \, \text{cm}, d = 5 \, \text{cm} $.

Substitute these into the equation:$\frac{1}{F} = \frac{1}{20} + \frac{-1}{10} – \frac{5}{(20)(-10)}$

Calculate each part:$\frac{1}{F} = \frac{1}{20} – \frac{2}{20} + \frac{5}{200}$

Simplifying further:$\frac{1}{F} = -\frac{1}{20} + \frac{1}{40} = \frac{-2 + 1}{40} = -\frac{1}{40}$

Therefore, $ F = -40 \, \text{cm} $. Since the focal length is negative, the system diverges light.

Notes:

• Lens Basics: Lenses are transparent materials that refract light. Two types: convex (converging) and concave (diverging).
• Lens Formula:
  $1f=1v−1u\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$ where
• f = focal length,
• v = image distance, and
• u = object distance.
• Sign Convention: Positive for distances in the direction of light; focal length depends on lens type.
• Power of Lens:
  P=100f(cm)P = \frac{100}{f(\text{cm})}P=f(cm)100​, measured in dioptres (D).
• Combined Lenses: Effective focal length:
  1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}
• Ray Diagrams: Aid in image formation analysis (magnification, nature, position).
• Complex Problems: Use stepwise substitution in formulas; consider system-wide impacts.

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Key Concept: Real vs Virtual Images

a) Real and inverted
[Solution Description]
In the context of lenses, particularly for concave lenses, the height of the image is considered positive when the image is virtual and erect. Conversely, it’s considered negative when the image is real and inverted.

Since we are dealing with a concave lens, which typically forms only virtual images for single-lens setups, a negative image height under the New Cartesian Sign Convention would be unusual. However, theoretically, it indicates that the image is real and inverted, though practically, concave lenses primarily form virtual, upright images.

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Key Concept: Angle of Refraction

c) $28.1^\circ$
[Solution Description]
To find the angle of refraction, we use Snell’s Law:
$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$
where $ n_1 = 1 $ (refractive index of air), $ \theta_1 = 45^\circ $ (angle of incidence), and $ n_2 = 1.5 $ (refractive index of glass). Solve for $ \theta_2 $:

$1 \times \sin(45^\circ) = 1.5 \times \sin(\theta_2)$

$\frac{\sqrt{2}}{2} = 1.5 \times \sin(\theta_2)$

$\sin(\theta_2) = \frac{\sqrt{2}}{3}$

Using inverse sine:

$\theta_2 \approx \arcsin\left(\frac{\sqrt{2}}{3}\right) \approx 28.1^\circ$
Thus, the angle of refraction inside the glass is approximately $28.1^\circ$.

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Key Concept: Basic Definition

b) $m = \frac{h’}{h}$
[Solution Description] The magnification (m) produced by a lens is defined as the ratio of the height of the image (h’) to the height of the object (h). Therefore, $m = \frac{h’}{h}$.

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Key Concept: Snell’s Law Application

c) 52°
[Solution Description]
To find the angle of refraction using Snell’s Law: $n_1 \sin i = n_2 \sin r$, where $n_1 = 1.5$, $i = 45^\circ$, and $n_2 = 1.33$.

Using this, we calculate $\sin r$:
$1.5 \times \sin 45^\circ = 1.33 \times \sin r.$

We know that $\sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707$.

Thus,
$1.5 \times 0.707 = 1.33 \times \sin r.$

Solving for $\sin r$, we get:
$\sin r = \frac{1.5 \times 0.707}{1.33} \approx 0.797.$

Taking the inverse sine of both sides gives:
$r \approx \arcsin(0.797) \approx 52.34^\circ.$

Therefore, the angle of refraction is approximately $52^\circ$.

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Key Concept: Advanced Reflection Concepts, Real-World Problem Solving

a) 25 cm behind the mirror
[Solution Description]
To find the position of the image formed by a convex mirror, we use the mirror formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
Given that the focal length $f$ is 20 cm and the object distance $u$ is 100 cm, note that for a convex mirror, both $f$ and $u$ are positive in this context. We substitute these values into the formula:
$\frac{1}{20} = \frac{1}{v} + \frac{1}{100}$
Rearranging for $\frac{1}{v}$ gives:
$\frac{1}{v} = \frac{1}{20} – \frac{1}{100}$
$\frac{1}{v} = \frac{5 – 1}{100} = \frac{4}{100} = \frac{1}{25}$

Thus, v = 25 cm. This means the image is formed 25 cm behind the mirror.

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Key Concept: Sign Convention Analysis

c) $-10$ cm, indicating a real image
[Solution Description]
Using the mirror formula:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
with $ u = -10 \, \text{cm} $ and $ f = -5 \, \text{cm} $, substitute:
$\frac{1}{v} + \frac{1}{(-10)} = \frac{1}{(-5)}$
Solve for $\frac{1}{v}$:
$\frac{1}{v} = \frac{1}{(-5)} + \frac{1}{10}$
Find a common denominator:
$\frac{1}{v} = \frac{-2}{10} + \frac{1}{10} = \frac{-1}{10}$
Thus, $ v = -10 \, \text{cm} $.

The negative sign indicates that the image is formed on the same side as the object, and it is real.

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Key Concept: Magnification Calculation

d) -2
[Solution Description]
The formula for magnification $m$ is given by $m = v/u$, where $v$ is the image distance and $u$ is the object distance. Given $v = +30$ cm and $u = -15$cm, we substitute these values into the formula:
$m = \frac{30}{-15} = -2.$
Hence, the magnification is -2. This indicates that the image is real and inverted with a size twice that of the object.

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Key Concept: Ray Diagram Analysis

c) Between the center of curvature and the focus

[Solution Description] For a concave mirror, when the object is placed beyond the center of curvature (C), the image formation follows these steps:

– A ray parallel to the principal axis reflects through the focus.

– A ray passing through the center of curvature reflects back on itself.

– The intersection of these reflected rays gives the image location.

In this scenario, the image forms between the center of curvature and the focus. Therefore, the image is real, inverted, and smaller than the object.

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Key Concept: Advanced Application, Problem Solving, Integration with Other Concepts

b) 16.67 cm, Real
[Solution Description]
To find the image distance $v$, we use the lens formula:
$\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$
Where $ f = 10 \, \text{cm} $ and $ u = -25 \, \text{cm} $.

Substituting these values into the lens formula:
$\frac{1}{10} = \frac{1}{v} + \frac{1}{25}$

Rearranging the equation to solve for $ \frac{1}{v} $:
$\frac{1}{v} = \frac{1}{10} – \frac{1}{25}$

Converting to a common denominator:
$\frac{1}{v} = \frac{5}{50} – \frac{2}{50} = \frac{3}{50}$

Therefore, $ v = \frac{50}{3} \approx 16.67 \, \text{cm} $.

Since the value of $ v $ is positive, the image is real and formed on the opposite side of the lens.

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Key Concept: Application of Sign Convention

b) Negative

[Solution Description] According to the New Cartesian Sign Convention, distances measured against the direction of the incident light are negative. Since the object is placed to the left of the mirror (opposite to the direction of incident light), the object distance u is negative. Therefore, the sign of the object distance is negative.

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Key Concept: Multi-Step Reasoning, Advanced Application

b) The image is real, inverted, and magnified.
[Solution Description]
To solve this problem, we first apply the mirror formula:

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

where $f$ is the focal length of the mirror, v is the image distance, and u is the object distance. Here, $f = -20$ cm and $u = -15$ cm (object distance is negative in the sign convention). Substituting these values into the formula:

$\frac{1}{-20} = \frac{1}{v} + \frac{1}{-15}$

Simplifying further:

$\frac{1}{v} = \frac{1}{-20} + \frac{1}{15}$

Calculating the right side:

$\frac{1}{v} = -\frac{3}{60} + \frac{4}{60} = \frac{1}{60}$

Therefore, v = 60 cm. Since v is positive, the image is real and inverted as it forms on the same side as the object.

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Key Concept: Mirror Type

c) Convex mirror
[Solution Description] A convex mirror always forms a virtual, erect, and diminished image irrespective of the object’s position. This property makes convex mirrors useful for applications like vehicle side mirrors.

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Key Concept: Complex Calculations, Real-world Problem Solving

b) Virtual, erect at 6.67 cm
[Solution Description]
We are given that the focal length $f$ of the concave mirror is -10 cm (since it is a concave mirror, the focal length is negative), and the object distance $u$ is -4 cm (the object is in front of the mirror). We need to find the image distance $v$.

Using the mirror formula:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

Substitute the known values:
$\frac{1}{v} – \frac{1}{4} = -\frac{1}{10}$

Simplifying,
$\frac{1}{v} = -\frac{1}{10} + \frac{1}{4}$

Convert to common denominator:
$\frac{1}{v} = -\frac{1}{10} + \frac{2.5}{10} = \frac{1.5}{10}$

Therefore,
$v = \frac{10}{1.5} = 6.67 \text{ cm}$

Since v is positive, the image is on the same side as the object, indicating it’s virtual and erect.

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