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Class 10 Physics – Chapter 14: Light

Board: Punjab Board
Subject: Physics
Class: 10th

A. Multiple Choice Questions (MCQs)

14.1 What type of image is always formed by a convex mirror?

(a) Real and inverted
(b) Virtual and diminished
(c) Real and magnified
(d) Virtual and magnified

Correct Answer: (b) Virtual and diminished

14.2 The optical device that uses a convex lens to magnify an object by creating a virtual image is:

(a) Camera
(b) Microscope
(c) Magnifying glass
(d) Slide projector

Correct Answer: (c) Magnifying glass

14.3 Which statement best describes the phenomenon of total internal reflection?

(a) Light bends when it enters a new medium.
(b) Light is completely reflected at the boundary of two media.
(c) Light slows down as it enters a denser medium.
(d) Light passes through a medium without changing direction.

Correct Answer: (b) Light is completely reflected at the boundary of two media.

14.4 A ray of light travelling from water to air bends:

(a) Towards the normal
(b) Away from the normal
(c) At \(90^\circ\) to the normal
(d) Parallel to the normal

Correct Answer: (b) Away from the normal

14.5 The optical device that works on the principle of inverting an image for projection is:

(a) Magnifying glass
(b) Photograph enlarger
(c) Slide projector
(d) Plane mirror

Correct Answer: (c) Slide projector

14.6 What happens to light when it enters an optical fibre?

(a) It is absorbed by the fibre.
(b) It undergoes total internal reflection.
(c) It gets refracted at every point.
(d) It travels in a straight line without deviation.

Correct Answer: (b) It undergoes total internal reflection.

14.7 Which phenomenon explains why a pencil appears bent when partially immersed in water?

(a) Reflection
(b) Diffraction
(c) Refraction
(d) Interference

Correct Answer: (c) Refraction

B. Short Answer Questions

14.1 What happens when light passes from air to water?

Answer:

When light passes from air (a rarer medium) into water (a denser medium), it slows down and bends towards the normal at the boundary between the two media due to refraction.

14.2 Define refractive index. What is the refractive index of air and water?

Answer:

Refractive Index (\(n\)): It is defined as the ratio of the speed of light in a vacuum (\(c\)) to the speed of light in a specific medium (\(v\)).

\[ n=\frac{c}{v} \]

Refractive index of air: Approximately \(1.00\)
Refractive index of water: \(1.33\)

14.3 What is the principle behind optical fibres?

Answer:

Optical fibres work on the principle of total internal reflection. Light signals entering the core strike the core-cladding interface at an angle greater than the critical angle and reflect repeatedly inside the core, allowing signals to travel along the fibre with minimal energy loss.

14.4 Describe briefly how a convex lens forms a real or virtual image.

Answer:

Real Image: When an object is placed beyond the principal focus (\(p>f\)), the light rays converge on the opposite side of the lens, forming a real, inverted image.
Virtual Image: When an object is placed within the focal length (\(p<f\)), the rays diverge after passing through the lens. When extended backwards, they appear to meet on the same side as the object, forming an upright, magnified, virtual image.

14.5 How does the power of a lens relate to its focal length?

Answer:

The power of a lens (\(P\)) is the reciprocal of its focal length (\(f\)), measured in metres.

\[ P=\frac{1}{f} \]

Its SI unit is the dioptre (D). A shorter focal length corresponds to a greater magnitude of optical power.

14.6 Write Snell’s law and define the terms used in the equation.

Answer:

Snell’s law is:

\[ n_1\sin i=n_2\sin r \]

Where:

\(n_1\) = Refractive index of the first medium
\(n_2\) = Refractive index of the second medium
\(i\) = Angle of incidence
\(r\) = Angle of refraction

For light travelling from air into a medium, the refractive index of the medium relative to air can be approximated by:

\[ n=\frac{\sin i}{\sin r} \]

C. Constructed Response Questions

14.1 When a pencil is partially dipped in water, it appears bent or broken. What principle of light explains this phenomenon, and how does it relate to changes in speed?

Answer:

This phenomenon is caused by refraction of light. Light travels faster in air than in water. When light rays from the submerged part of the pencil pass from water into air, they speed up and bend away from the normal. The human eye traces these rays backwards, making the submerged part appear shifted from its actual position. As a result, the pencil appears bent at the water surface.

14.2 Why can a prism separate white light into its component colours while a plane glass slab cannot, even though both cause refraction?

Answer:

Glass Prism: The non-parallel refracting faces of a prism cause different wavelengths (colours) of light to bend by different amounts. This separates white light into its component colours, producing a spectrum.
Glass Slab: The opposite faces of a glass slab are parallel. The light rays bend at the first surface and bend in the opposite direction at the second surface. The emergent rays are parallel to the incident rays but laterally displaced, so the colours are not separated into a visible spectrum.

14.3 Why does light undergo total internal reflection when travelling from water to air at a certain angle but not the other way around?

Answer:

Total internal reflection occurs only when light travels from an optically denser medium to an optically rarer medium.

When light travels from water (denser) to air (rarer), it bends away from the normal. If the angle of incidence exceeds the critical angle, the light is reflected completely back into the water.

When light travels from air to water, it bends towards the normal. Therefore, total internal reflection cannot occur in this direction.

14.4 Optical fibres use total internal reflection for signal transmission. What would happen if the cladding had a higher refractive index than the core, and why?

Answer:

If the cladding had a higher refractive index than the core, total internal reflection would not occur at the core-cladding boundary. Instead, light could refract into the cladding rather than remaining confined within the core. This would cause significant signal loss and interfere with transmission.

14.5 A person uses a magnifying glass to read small print. What must be the relationship between the object distance and the focal length of the lens for the image to be upright and magnified?

Answer:

The object must be placed closer to the convex lens than its focal length. In other words, the object distance (\(p\)) must be less than the focal length (\(f\)):

\[ p<f \]

This produces an upright, magnified, and virtual image.

D. Comprehensive Questions

14.1 Explain the terms: normal, angle of incidence, and angle of reflection in the context of light reflection. Provide a diagram to illustrate these terms.

Answer:

Normal: An imaginary line drawn perpendicular to the reflecting surface at the point of incidence.
Angle of Incidence (\(\theta_i\)): The angle between the incident ray and the normal.
Angle of Reflection (\(\theta_r\)): The angle between the reflected ray and the normal.

Diagram:

Angle of incidence and angle of reflection are measured from the normal.

According to the law of reflection:

\[ \theta_i=\theta_r \]

14.2 Define the terms critical angle and total internal reflection. Explain the conditions required for total internal reflection to occur.

Answer:

Critical Angle (\(C\)): The angle of incidence in an optically denser medium for which the angle of refraction in the optically rarer medium is \(90^\circ\).

Total Internal Reflection: The complete reflection of light back into the denser medium when the angle of incidence exceeds the critical angle.

Conditions for total internal reflection:

Light must travel from an optically denser medium to an optically rarer medium.
The angle of incidence must be greater than the critical angle:

\[ i>C \]

14.3 Derive the equation \(n=\frac{1}{\sin C}\) for the refractive index in terms of the critical angle \(C\). Include all steps and assumptions.

Answer:

According to Snell’s law:

\[ n_1\sin i=n_2\sin r \]

Assumptions:

Medium 1 is the denser medium, with refractive index \(n_1=n\).
Medium 2 is air, with refractive index \(n_2\approx1\).
At the critical angle, \(i=C\) and \(r=90^\circ\).

Derivation:

Substituting these values into Snell’s law:

\[ n\sin C=1\times\sin90^\circ \]

Since:

\[ \sin90^\circ=1 \]

Therefore:

\[ n\sin C=1 \]

Dividing both sides by \(\sin C\):

\[ \boxed{n=\frac{1}{\sin C}} \]

This equation gives the refractive index of a medium relative to air in terms of its critical angle.

14.4 Describe the structure of an optical fibre. Write any two advantages.

Answer:

Structure of an optical fibre:

Core: The inner central part made of high-quality glass or plastic. It has a higher refractive index (\(n_1\)).
Cladding: The outer glass or plastic layer surrounding the core. Its refractive index is lower than that of the core (\(n_2<n_1\)).
Protective Coating: The outer protective layer that shields the fibre from moisture and physical damage.

Advantages:

High bandwidth, allowing large volumes of data to be transmitted per second.
Immunity to electromagnetic interference and minimal signal loss over long distances.

14.5 What is long-sightedness? Discuss its causes.

Answer:

Long-sightedness (Hypermetropia): A vision defect in which distant objects can generally be seen clearly, but nearby objects appear blurred because light from nearby objects is focused behind the retina.

Causes:

The eyeball is too short from front to back.
The focal length of the eye lens is too long, meaning its converging power is insufficient.

Correction: Long-sightedness is corrected using a convex lens.

E. Numerical Problems

Problem 14.1

Given:

Speed of light in water: \(v=2.25\times10^8\ \text{m s}^{-1}\)
Speed of light in vacuum: \(c=3.0\times10^8\ \text{m s}^{-1}\)

To Find:

Refractive index of water, \(n=?\)

Formula:

\[ n=\frac{c}{v} \]

Calculation:

\[ n=\frac{3.0\times10^8}{2.25\times10^8} \]

\[ n=1.33 \]

Answer: The refractive index of water is \(1.33\).

Problem 14.2

Given:

Refractive index of air: \(n_1=1.0\)
Refractive index of glass: \(n_2=1.5\)
Angle of incidence: \(i=30^\circ\)

To Find:

Angle of refraction, \(r=?\)

Formula:

\[ n_1\sin i=n_2\sin r \]

Calculation:

\[ 1.0\sin30^\circ=1.5\sin r \]

Since \(\sin30^\circ=0.5\):

\[ 1.0\times0.5=1.5\sin r \]

\[ \sin r=\frac{0.5}{1.5}=0.3333 \]

Taking the inverse sine:

\[ r=\sin^{-1}(0.3333) \]

\[ r\approx19.47^\circ\approx19.5^\circ \]

Answer: The angle of refraction is approximately \(19.5^\circ\).

Problem 14.3

Given:

Refractive index of air: \(n_1=1.0\)
Angle of incidence: \(i=40^\circ\)
Refractive index of glass: \(n_2=1.52\)

To Find:

Angle of refraction, \(r=?\)

Formula:

\[ n_1\sin i=n_2\sin r \]

Calculation:

\[ 1.0\sin40^\circ=1.52\sin r \]

Using \(\sin40^\circ\approx0.6428\):

\[ 0.6428=1.52\sin r \]

\[ \sin r=\frac{0.6428}{1.52} \]

\[ \sin r\approx0.4229 \]

Taking the inverse sine:

\[ r=\sin^{-1}(0.4229) \]

\[ r\approx25^\circ \]

Answer: The angle of refraction is approximately \(25^\circ\).

Problem 14.4

Given:

Angle of incidence: \(i=45^\circ\)
Angle of refraction: \(r=30^\circ\)

To Find:

Refractive index of the liquid, \(n=?\)

Formula:

\[ n=\frac{\sin i}{\sin r} \]

Calculation:

\[ n=\frac{\sin45^\circ}{\sin30^\circ} \]

\[ n=\frac{0.7071}{0.5} \]

\[ n=1.414\approx1.41 \]

Answer: The refractive index of the liquid is approximately \(1.41\).

Problem 14.5

Given:

Angle of incidence: \(i=50^\circ\)
Angle of refraction: \(r=32^\circ\)
Speed of light in vacuum: \(c=3.0\times10^8\ \text{m s}^{-1}\)

To Find:

(a) Refractive index of the liquid, \(n=?\)

(b) Speed of light in the liquid, \(v=?\)

Formulae:

\[ n=\frac{\sin i}{\sin r} \]

\[ v=\frac{c}{n} \]

Calculation:

(a) Refractive index

\[ n=\frac{\sin50^\circ}{\sin32^\circ} \]

\[ n=\frac{0.7660}{0.5299} \]

\[ n\approx1.445\approx1.45 \]

(b) Speed of light in the liquid

Using the unrounded refractive index:

\[ v=\frac{3.0\times10^8}{1.445} \]

\[ v\approx2.08\times10^8\ \text{m s}^{-1} \]

Answer:

(a) Refractive index \(=1.45\)

(b) Speed of light in the liquid \(\approx2.08\times10^8\ \text{m s}^{-1}\)

Problem 14.6

Given:

– Refractive index of glass: \(n=1.6\)

To Find:

Critical angle, \(C=?\)

Formula:

\[ \sin C=\frac{1}{n} \]

Calculation:

\[ \sin C=\frac{1}{1.6}=0.625 \]

Taking the inverse sine:

\[ C=\sin^{-1}(0.625) \]

\[ C\approx38.68^\circ \]

Answer: The critical angle is approximately \(38.68^\circ\).

Problem 14.7

Given:

Object distance: \(p=+15.0\ \text{cm}\)
Image distance: \(q=-7.5\ \text{cm}\)

To Find:

Focal length, \(f=?\)

Formula:

\[ \frac{1}{f}=\frac{1}{p}+\frac{1}{q} \]

Calculation:

\[ \frac{1}{f}=\frac{1}{15.0}+\frac{1}{-7.5} \]

\[ \frac{1}{f}=\frac{1}{15.0}-\frac{1}{7.5} \]

\[ \frac{1}{f}=\frac{1-2}{15.0}=-\frac{1}{15.0} \]

Therefore:

\[ f=-15.0\ \text{cm} \]

Answer: The focal length is \(-15\ \text{cm}\).

Problem 14.8

Given:

Object height: \(h_o=30\ \text{cm}\)
Object distance: \(p=+10.5\ \text{cm}\)
Focal length of concave mirror: \(f=+16\ \text{cm}\)

To Find:

(a) Image distance, \(q=?\)

(b) Image height, \(h_i=?\)

(a) Calculate Image Distance

Formula:

\[ \frac{1}{f}=\frac{1}{p}+\frac{1}{q} \]

Rearranging:

\[ \frac{1}{q}=\frac{1}{f}-\frac{1}{p} \]

Calculation:

\[ \frac{1}{q}=\frac{1}{16}-\frac{1}{10.5} \]

\[ \frac{1}{q}=0.0625-0.09524 \]

\[ \frac{1}{q}\approx-0.03274 \]

\[ q\approx-30.54\ \text{cm} \]

Therefore:

\[ q\approx-30.6\ \text{cm} \]

(b) Calculate Image Height

Using the magnification relationship provided in the source:

\[ h_i=h_o\times\frac{|q|}{p} \]

Substituting the values:

\[ h_i=30\times\frac{30.54}{10.5} \]

\[ h_i\approx87.26\ \text{cm} \]

\[ h_i\approx87.3\ \text{cm} \]

Answer:

(a) Image distance \(=-30.6\ \text{cm}\), indicating a virtual image under the sign convention used.

(b) Image height \(=87.3\ \text{cm}\) in magnitude.

Problem 14.9

Given:

Object distance: \(p=+30\ \text{cm}\)
Focal length of concave lens: \(f=-15\ \text{cm}\)

To Find:

Image distance, \(q=?\)
Nature of the image

Formula:

\[ \frac{1}{f}=\frac{1}{p}+\frac{1}{q} \]

Calculation:

\[ \frac{1}{-15}=\frac{1}{30}+\frac{1}{q} \]

Rearranging:

\[ \frac{1}{q}=-\frac{1}{15}-\frac{1}{30} \]

\[ \frac{1}{q}=\frac{-2-1}{30} \]

\[ \frac{1}{q}=-\frac{3}{30}=-\frac{1}{10} \]

Therefore:

\[ q=-10\ \text{cm} \]

Answer:

Image distance: \(-10\ \text{cm}\)
Nature of image: Virtual, upright, and formed on the same side as the object.

Note: The source mentions an alternative answer of \(-30\ \text{cm}\), but substituting the supplied values into the stated lens formula gives \(-10\ \text{cm}\).

Notes by Sir Dilawar

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