11th physics Optics – Complete Chapter Solutions

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11th physics Optics – Complete Chapter Solutions

Published: Jan 06, 2026 •

Q.1 Choose the correct option.

Q(i). As per recent understanding, light consists of

(A) rays
(B) waves
(C) corpuscles
(D) photons obeying the rules of waves

Answer:

Correct option: (D) photons obeying the rules of waves

Explanation:

According to modern quantum theory, light shows dual nature. It behaves as particles called photons and also exhibits wave properties like interference and diffraction.

Q(ii). Consider the optically denser lenses P, Q, R and S drawn below. According to Cartesian sign convention, which of these have positive focal length?

(A) Only P
(B) Only P and Q
(C) Only P and R
(D) Only Q and S

Answer:

Correct option: (B) Only P and Q

Explanation:

According to the Cartesian sign convention:

Focal length is positive for converging (convex) lenses
Focal length is negative for diverging (concave) lenses

Since the lenses are made of optically denser material:

P → Convex lens → Positive focal length
Q → Convex lens → Positive focal length
R → Concave lens → Negative focal length
S → Concave lens → Negative focal length

Hence, lenses having positive focal length are:

P and Q only

Q(iii). Two plane mirrors are inclined at an angle of 40° between them. Number of images formed is

(A) Only 8
(B) Only 9
(C) 8 or 9
(D) 9 or 10

Answer:

Correct option: (C) 8 or 9

Explanation:

Number of images formed by two inclined mirrors:

$$ n = \frac{360^\circ}{\theta} $$

Here,

$$ n = \frac{360}{40} = 9 $$

Since 9 is an integer, number of images may be:

$$ n = 8 \text{ or } 9 $$

Q(iv). A concave mirror of radius of curvature 40 cm produces an image twice the size of the object for shaving. Object distance is

(A) 10 cm only
(B) 20 cm only
(C) 30 cm only
(D) 10 cm or 30 cm

Answer:

Correct option: (D) 10 cm or 30 cm

Explanation:

Radius of curvature:

$$ R = 40 \text{ cm} \Rightarrow f = \frac{R}{2} = 20 \text{ cm} $$

Magnification:

$$ m = \pm 2 $$

Using:

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

Solving mirror formula gives:

$$ u = 10 \text{ cm or } 30 \text{ cm} $$

Q(v). Which aberration does NOT occur in spherical mirrors?

(A) Chromatic aberration
(B) Coma
(C) Distortion
(D) Spherical aberration

Answer:

Correct option: (A) Chromatic aberration

Explanation:

Chromatic aberration occurs due to dispersion in lenses. Mirrors reflect all wavelengths equally, so chromatic aberration does not occur.

Q(vi). A monkey sees a fish at 2.5 m and fish sees monkey at 1.6 m. Actual distances are equal. Refractive index of water is

(A) 6/5
(B) 5/4
(C) 4/3
(D) 7/5

Answer:

Correct option: (B) 5/4

Explanation:

Refractive index:

$$ \mu = \frac{\text{real depth}}{\text{apparent depth}} $$

Given:

$$ \mu = \frac{2.5}{2.0} = 1.25 = \frac{5}{4} $$

Q(vii). Total internal reflection is involved in

P) Mirage
Q) Rainbow
R) Optical fibre
S) Glittering of diamond

(A) Only R and S
(B) Only R
(C) Only P, R and S
(D) All the four

Answer:

Correct option: (C) Only P, R and S

Explanation:

Mirage, optical fibre and diamond brilliance involve total internal reflection. Rainbow involves refraction and internal reflection, not TIR.

Q(viii). A person using spectacles of power −2 D uses lenses which are

(A) bi-concave
(B) double concave
(C) concavo-convex
(D) convexo-concave

Answer:

Correct option: (A) bi-concave

Explanation:

Negative power indicates a concave lens. Spectacles generally use bi-concave lenses for myopia.

Q(ix). A spherical marble (μ = 1.5, radius = 1.5 cm) has an air bubble at centre. Its apparent position is

(A) 1 cm inside
(B) at centre
(C) 5/3 cm inside
(D) 2 cm inside

Answer:

Correct option: (B) at centre

Explanation:

Due to symmetry of refraction at spherical surfaces, the bubble at centre appears at the centre.

Q(x). Select the WRONG statement.

(A) Smaller prism angle gives greater dispersion
(B) Right-angled prism is used for TIR
(C) Deviation is constant for thin prism
(D) Minimum incidence is needed for emergence

Answer:

Correct option: (A)

Explanation:

Angular dispersion depends on material, not prism angle. Hence statement (A) is wrong.

Q(xi). Prism with maximum dispersive power is

(A) 7q, 10q
(B) 8q, 11q
(C) 12q, 16q
(D) 10q, 14q

Answer:

Correct option: (C)

Explanation:

Dispersive power depends on difference:

$$ \Delta \delta = \delta_v - \delta_r $$

Maximum difference:

$$ 16q - 12q = 4q $$

Q(xii). Which is NOT involved in rainbow formation?

(A) Refraction
(B) Angular dispersion
(C) Angular deviation
(D) Total internal reflection

Answer:

Correct option: (C) Angular deviation

Explanation:

Rainbow formation involves refraction, dispersion and internal reflection. Angular deviation is not a separate phenomenon here.

Q(xiii). Correct statements for a simple microscope are

P) Object is kept within least distance of vision
Q) Image is biggest when object is at focus
R) It is a convex lens

(A) Only P
(B) P and Q
(C) Q and R
(D) P and R

Answer:

Correct option: (D) P and R

Explanation:

Maximum magnification occurs when image is at near point, not at focus. Simple microscope is a convex lens.

2. Answer the following questions.

Question 2 (i) As per recent development, what is the nature of light? Wave optics and particle nature of light are used to explain which phenomena of light respectively?

Answer:

According to modern physics, light exhibits a dual nature, i.e., it behaves both as a wave and as a particle.

Wave Nature of Light:
Wave optics explains phenomena that arise due to the wave behavior of light such as:

Interference
Diffraction
Polarization

These phenomena occur due to superposition and transverse nature of light waves.

Particle Nature of Light:
Particle nature of light explains phenomena where light behaves as discrete packets of energy called photons, such as:

Photoelectric effect
Compton effect
Black body radiation

Hence, light has a dual nature and both wave optics and particle nature are required to explain different optical phenomena.

Question 2 (ii) Which phenomena can be satisfactorily explained using ray optics? State the assumptions on which ray optics is based.

Answer:

Ray optics can satisfactorily explain the following phenomena:

Reflection of light
Refraction of light
Formation of images by mirrors and lenses
Dispersion of light by prisms

Ray optics fails to explain interference, diffraction, and polarization.

Assumptions of Ray Optics:

Light travels in straight lines called rays.
The wavelength of light is very small compared to the size of optical components.
The path of light can be represented by geometrical lines.
Effects due to the wave nature of light are neglected.

These assumptions are valid when the dimensions of apertures and obstacles are much larger than the wavelength of light.

Question 2 (iii) What is focal power of a spherical mirror or a lens? What may be the reason for using $ \dfrac{1}{f} $ as its expression?

Answer:

The focal power of a spherical mirror or a lens is defined as the reciprocal of its focal length.

Mathematically,

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

where,

$ P $ = focal power (in dioptre)
$ f $ = focal length (in metre)

Unit of Power:
The SI unit of power is dioptre (D).

Reason for using $ \dfrac{1}{f} $:

Power gives a direct measure of the converging or diverging ability of a lens or mirror.
A smaller focal length means greater bending of light and hence larger power.
Using reciprocal of focal length makes comparison between optical devices easier.

Thus, focal power is expressed as $ \dfrac{1}{f} $ for convenience and practical usefulness.

Question (iv)> At which positions of the object do spherical mirrors produce:
(i) diminished image,
(ii) magnified image?

Answer:

Step 1: Understanding spherical mirrors

Spherical mirrors are of two types:

Concave mirror – converging mirror
Convex mirror – diverging mirror

Step 2: Diminished image

A diminished image is smaller than the object.

Concave mirror: When the object is placed beyond the centre of curvature (C), the image is formed between $F$ and $C$ , and is diminished.
Convex mirror: For all positions of the object in front of a convex mirror, the image formed is always diminished.

Step 3: Magnified image

A magnified image is larger than the object.

Concave mirror:
- When the object is placed between the centre of curvature (C) and focus (F)
- When the object is placed between the focus (F) and the pole (P)
Convex mirror: Convex mirrors never produce magnified images.

Question (v) State the restrictions for having images produced by spherical mirrors to be appreciably clear.

Answer:

Step 1: Meaning of appreciably clear image

An appreciably clear image is one that is sharp, bright, and well-defined.

Step 2: Restrictions / Conditions

For images formed by spherical mirrors to be appreciably clear, the following conditions must be satisfied:

The mirror should have a smooth and well-polished reflecting surface.
The object should be placed on the principal axis of the mirror.
The size of the mirror aperture should be small to reduce spherical aberration.
Light rays used should be paraxial rays (rays close to the principal axis).
The mirror should be free from dust, scratches, or distortions.

Conclusion:

Only when these restrictions are satisfied does a spherical mirror form a sharp and clearly visible image.

Question (vi) Explain spherical aberration for spherical mirrors. How can it be minimized? Can it be eliminated by some curved mirrors?

Answer:

Meaning of Spherical Aberration:

Spherical aberration is a defect of spherical mirrors in which rays of light parallel to the principal axis do not converge (or appear to diverge) at a single point after reflection.

For a concave spherical mirror:

Rays close to the principal axis (paraxial rays) focus at a point F.
Rays far from the principal axis (marginal rays) focus at a point closer to the mirror.

As a result, a sharp image is not formed and the image appears blurred.

Reason:

This happens because the mirror is spherical in shape, and different zones of the mirror have different focal lengths.

Minimization of Spherical Aberration:

By using only paraxial rays (placing an aperture stop to block marginal rays).
By reducing the aperture of the mirror.

Elimination of Spherical Aberration:

Yes, spherical aberration can be completely eliminated by using a parabolic mirror. A parabolic mirror reflects all rays parallel to the principal axis to a single focus.

Question (vii) Define absolute refractive index and relative refractive index. Explain in brief with an illustration for each.

Answer:

(a) Absolute Refractive Index:

The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in that medium.

Mathematically,

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

where,

$c$ = speed of light in vacuum
$v$ = speed of light in the medium

Since $v < c$ , the value of $n > 1$ .

Illustration:

Light entering glass from vacuum slows down, hence glass has an absolute refractive index greater than unity.

(b) Relative Refractive Index:

The relative refractive index of one medium with respect to another is defined as the ratio of the speed of light in the first medium to the speed of light in the second medium.

Mathematically,

$n_{21} = \frac{v_1}{v_2} = \frac{n_2}{n_1}$

where,

$$v_1$$ , $$v_2$$ are speeds of light in medium 1 and medium 2 respectively
$$n_1$$ , $$n_2$$ are their absolute refractive indices

Illustration:

When light travels from water to glass, the refractive index of glass with respect to water is called the relative refractive index.

Question (viii) Explain ‘mirage’ as an illustration of refraction.

Answer:

A mirage is an optical illusion observed on hot summer days, especially on highways or in deserts, where a pool of water appears on the ground but disappears when approached. This phenomenon occurs due to refraction of light in air of varying densities.

Step-by-step explanation:

1. On a hot day, the air near the ground becomes very hot and hence less dense, while the air at higher levels is relatively cooler and denser.

2. Therefore, the refractive index of air decreases gradually from top to bottom.

3. Light rays coming from a distant object (like the sky or a tree) travel from denser air to rarer air and undergo continuous refraction.

4. At a certain height near the ground, the angle of incidence becomes equal to the critical angle, and the ray undergoes total internal reflection.

5. These reflected rays enter the observer’s eyes, and the brain assumes them to come from the ground, forming a virtual inverted image of the sky.

6. This image looks like water on the road and is called a mirage.

Hence, a mirage is an excellent illustration of refraction and total internal reflection occurring in the atmosphere.

Question (ix) Under what conditions is total internal reflection possible? Explain it with a suitable example. Define critical angle of incidence and obtain an expression for it.

Answer:

Total Internal Reflection (TIR) is the phenomenon in which a light ray is completely reflected back into the denser medium when it tries to pass from a denser medium to a rarer medium.

Conditions for Total Internal Reflection:

1. Light must travel from an optically denser medium to an optically rarer medium.

2. The angle of incidence in the denser medium must be greater than the critical angle.

If any one of these conditions is not satisfied, total internal reflection will not occur.

Example of Total Internal Reflection:

When light travels from water to air or glass to air at an angle greater than the critical angle, it gets totally reflected back into water or glass. This principle is used in optical fibres for communication.

Critical Angle of Incidence:

The critical angle (C) is defined as the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is $90^\circ$ .

Derivation of Expression for Critical Angle:

Let a ray of light travel from a denser medium (refractive index $\mu_1$ ) to a rarer medium (refractive index $\mu_2$ ).

At the critical angle, angle of incidence $$i = C$$ and angle of refraction $r = 90^\circ$ .

Using Snell’s law:

\mu_1 \sin C = \mu_2 \sin 90^\circ

Since $\sin 90^\circ = 1$ ,

\mu_1 \sin C = \mu_2

Hence,

\sin C = \frac{\mu_2}{\mu_1}

This is the required expression for the $critical angle$.

Thus, the critical angle depends on the refractive indices of the two media.

Question (x) Describe construction and working of an optical fibre. What are the advantages of optical fibre communication over electronic communication?

Answer:

Optical Fibre:

An optical fibre is a thin, flexible, transparent dielectric medium used to transmit light signals over long distances using the principle of Total Internal Reflection (TIR).

Construction of Optical Fibre:

An optical fibre consists of the following three main parts:

Core:
It is the central cylindrical region made of high refractive index glass or plastic. Light signals propagate through the core.
Cladding:
It surrounds the core and is made of glass with refractive index slightly less than that of the core. It enables total internal reflection of light inside the core.
Protective Jacket:
It is the outer covering made of plastic that protects the fibre from mechanical damage and moisture.

Working of Optical Fibre:

When a light ray enters the core of the optical fibre at an angle greater than the critical angle, it undergoes total internal reflection at the core–cladding interface.

Due to repeated total internal reflections, the light signal travels through the fibre with very little loss of energy and reaches the other end of the fibre.

Thus, information in the form of light pulses can be transmitted over long distances efficiently.

Advantages of Optical Fibre Communication:

Very high bandwidth and data transmission capacity.
Low signal loss, suitable for long-distance communication.
Immune to electromagnetic interference.
Smaller size and lighter weight.
High security as signals are difficult to tap.

Question (xi) Why is a prism binoculars preferred over traditional binoculars? Describe its working in brief.

Answer:

Reason for Preference of Prism Binoculars:

Prism binoculars are preferred over traditional (astronomical) binoculars because they provide an erect and laterally correct image while being more compact and comfortable to use.

Working of Prism Binoculars:

A prism binocular consists of two identical telescopes placed side by side, each containing a pair of right-angled prisms.

Light from a distant object enters through the objective lens and forms an inverted image. This light then passes through the prism system.

The prisms cause multiple total internal reflections, which:

Erect the inverted image
Correct lateral inversion
Fold the light path, reducing the length of the binocular

Finally, the eyepiece magnifies the erect image for observation by the eye.

Advantages of Prism Binoculars:

Compact and lightweight design
Erect and laterally correct image
Higher brightness and clarity
Convenient for terrestrial observations

xii) A spherical surface separates two transparent media. Derive an expression that relates object and image distances with the radius of curvature for a point object. Clearly state the assumptions.

Answer:

Consider a spherical refracting surface of radius of curvature $$R$$ separating two transparent media. Let the refractive index of the first medium (object side) be $$n_1$$ and that of the second medium (image side) be $$n_2$$ .

A point object $$O$$ is placed on the principal axis at a distance $$u$$ from the pole $$P$$ of the spherical surface. After refraction, an image $$I$$ is formed at a distance $$v$$ from the pole. Let $$C$$ be the centre of curvature of the surface.

Step 1: Geometry of refraction

Consider a paraxial ray from object $$O$$ incident at point $$A$$ on the spherical surface. Let the angles made by the ray with the normal at point $$A$$ be:

Angle of incidence $$i$$ and angle of refraction $$r$$ .

Using small angle approximation (paraxial approximation),

i \approx \angle OAC = \alpha + \gamma

r \approx \angle IAC = \beta + \gamma

where,

\alpha \approx \frac{h}{u}$, $\quad\beta \approx \frac{h}{v}$, $\quad\gamma \approx \frac{h}{R}

Here $$h$$ is the height of the incident ray from the principal axis.

Step 2: Apply Snell's Law

According to Snell’s law,

n_1 \sin i = n_2 \sin r

For small angles, $\sin i \approx i$ and $\sin r \approx r$ .

n_1 (\alpha + \gamma) = n_2 (\beta + \gamma)

Step 3: Substitute values

n_1 \left( \frac{h}{u} + \frac{h}{R} \right) = n_2 \left( \frac{h}{v} + \frac{h}{R} \right)

Cancel $$h$$ from both sides:

n_1 \left( \frac{1}{u} + \frac{1}{R} \right) = n_2 \left( \frac{1}{v} + \frac{1}{R} \right)

Step 4: Rearranging the equation

\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}

This is the required relation between object distance, image distance, and radius of curvature.

Final Formula (Refraction at a Spherical Surface):

\boxed{ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} }

Assumptions:

Rays are paraxial (small angle approximation is valid).
The spherical surface is smooth and of small aperture.
Refractive indices $$n_1$$ and $$n_2$$ are uniform.
Distances are measured from the pole using sign convention.

Question (xiii) Derive lens makers’ equation. Why is it called so? Under which conditions focal length $f$ and radii of curvature $R$ are numerically equal for a lens?

Answer:

Lens Maker’s Equation:

The lens maker’s equation gives the relation between the focal length of a lens, the refractive index of the lens material, and the radii of curvature of its two surfaces.

Consider a thin lens of refractive index $n$ placed in air. Let

$$R_1$$ = radius of curvature of the first surface
$$R_2$$ = radius of curvature of the second surface
$$f$$ = focal length of the lens

Refraction at the first surface:

Using the refraction formula at a spherical surface,

\frac{n}{v_1} - \frac{1}{u} = \frac{n - 1}{R_1}

For an object at infinity, $u = \infty$ , hence

\frac{n}{v_1} = \frac{n - 1}{R_1}

Refraction at the second surface:

The image formed by the first surface acts as the object for the second surface. Applying the refraction formula,

\frac{1}{v} - \frac{n}{v_1} = \frac{1 - n}{R_2}

Substituting the value of $\frac{n}{v_1}$ from the first equation:

\frac{1}{v} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

For an object at infinity, the final image is formed at the focus, so $$v = f$$ .

\boxed{\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}

This is the lens maker’s equation.

Why is it called $Lens Maker’s Equation$?

It is called the lens maker’s equation because it helps lens manufacturers decide the required radii of curvature of the lens surfaces for a desired focal length and refractive index of the material.

Condition for $$f$$ and $$R$$ to be numerically equal:

For a thin symmetric lens,

R_1 = R \quad \text{and} \quad R_2 = -R

Substituting in the lens maker’s equation:

\frac{1}{f} = (n - 1)\left(\frac{1}{R} - \left(-\frac{1}{R}\right)\right)

\frac{1}{f} = \frac{2(n - 1)}{R}

For $$f = R$$ numerically,

$2(n - 1) = 1$

$n = 1.5$

Hence, the focal length and radius of curvature are numerically equal when:

The lens is thin and symmetric
The lens is placed in air
The refractive index of the lens material is $$n = 1.5$$

2. Answer the following questions in detail.

Question 2 (i) What are different types of dispersion of light? Why do they occur?

Answer:

Dispersion of light is the phenomenon of splitting of white light into its constituent colours (VIBGYOR) when it passes through a refracting medium such as a prism.

Types of Dispersion of Light:

1. Normal Dispersion

In normal dispersion, the refractive index of the medium decreases with increase in wavelength. Hence, violet light deviates the most and red light deviates the least.

Mathematically,

\mu_v > \mu_r

This type of dispersion is commonly observed in glass prisms for visible light.

2. Anomalous Dispersion

In anomalous dispersion, the refractive index increases with increase in wavelength over a certain range. This behaviour is observed near absorption bands of the medium.

Mathematically,

\mu_r > \mu_v

Why does dispersion occur?

Dispersion occurs because the refractive index of a material depends on the wavelength of light. Different colours have different wavelengths, and hence they travel with different speeds in a medium. This results in different angles of refraction for different colours, causing separation.

Question 2 (ii) Define angular dispersion for a prism. Obtain its expression for a thin prism. Relate it with the refractive indices of the material of the prism for corresponding colours.

Answer:

Angular Dispersion:

Angular dispersion of a prism is defined as the angular separation between the rays of two extreme colours (usually red and violet) emerging from the prism.

If $\delta_v$ is the angle of deviation for violet light and $\delta_r$ is the angle of deviation for red light, then

\text{Angular dispersion} = \delta_v - \delta_r

Expression for Angular Dispersion of a Thin Prism:

For a thin prism, the angle of deviation is given by:

\delta = (\mu - 1)A

where,
$\mu$ = refractive index of the prism material
$$A$$ = angle of the prism

For violet light:

\delta_v = (\mu_v - 1)A

For red light:

\delta_r = (\mu_r - 1)A

Angular dispersion is:

\delta_v - \delta_r = [(\mu_v - 1) - (\mu_r - 1)]A

\delta_v - \delta_r = (\mu_v - \mu_r)A

Relation with Refractive Indices:

Thus, angular dispersion depends on the difference between the refractive indices of the prism material for violet and red light and the angle of the prism.

Hence,

\text{Angular dispersion} \propto (\mu_v - \mu_r)

A material with a larger difference in refractive indices for different colours produces greater dispersion.

Question (iii) Explain and define dispersive power of a transparent material. Obtain its expressions in terms of (i) angles of deviation and (ii) refractive indices.

Answer:

Dispersion of Light:

When white light passes through a transparent medium like a prism, it splits into its constituent colours. This phenomenon is called dispersion of light.

Definition of Dispersive Power:

The dispersive power of a transparent material is defined as the ratio of the angular dispersion between two extreme colours (usually violet and red) to the mean deviation produced by the prism.

Expression of Dispersive Power in terms of Angles of Deviation:

Let:

$\delta_v$ = angle of deviation for violet light
$\delta_r$ = angle of deviation for red light
$\delta_y$ = angle of deviation for mean (yellow) light

Angular dispersion is given by:

\[ \delta_v - \delta_r \]

Mean deviation is given by:

\[ \delta_y \]

Hence, the dispersive power $\omega$ is:

\[ \omega = \frac{\delta_v - \delta_r}{\delta_y} \]

Expression of Dispersive Power in terms of Refractive Indices:

Let:

$\mu_v$ = refractive index for violet light
$\mu_r$ = refractive index for red light
$\mu_y$ = refractive index for yellow (mean) light

For a thin prism, the angle of deviation $\delta$ is given by:

\[ \delta = (\mu - 1)A \]

where $A$ is the angle of the prism.

Angular dispersion:

\[ \delta_v - \delta_r = (\mu_v - \mu_r)A \]

Mean deviation:

\[ \delta_y = (\mu_y - 1)A \]

Substituting in the definition of dispersive power:

\[ \omega = \frac{(\mu_v - \mu_r)A}{(\mu_y - 1)A} \]

Cancelling $A$:

\[ \boxed{\omega = \frac{\mu_v - \mu_r}{\mu_y - 1}} \]

Conclusion:

Dispersive power depends only on the nature of the material and not on the angle of the prism. Materials with higher dispersive power produce greater colour separation.

Question (iv)

(i) State the conditions under which a rainbow can be seen.

Answer:

A rainbow can be observed only when the following conditions are satisfied:

The Sun must be shining and should be behind the observer.
There must be water droplets (rain, mist, or spray) present in front of the observer.
The Sun should be at a low altitude (usually in the morning or evening).
The observer’s eye, the water droplets, and the Sun must lie in the same vertical plane.

Under these conditions, sunlight undergoes refraction, dispersion, and internal reflection inside water droplets, leading to the formation of a rainbow.

(ii) Explain the formation of a primary rainbow. For which angular range with the horizontal is it visible?

Answer:

The primary rainbow is formed due to the interaction of sunlight with spherical water droplets present in the atmosphere.

Step-by-step explanation:

Refraction and dispersion:
When white sunlight enters a raindrop, it gets refracted and dispersed into its constituent colours due to different wavelengths.
Internal reflection:
The dispersed light undergoes one total internal reflection at the back surface of the raindrop.
Refraction on emergence:
The light refracts again while emerging out of the raindrop.

After these processes, different colours emerge at different angles. Red light deviates the least, while violet light deviates the most.

The angle of deviation for red light is approximately $42^\circ$ and for violet light is approximately $40^\circ$ .

Hence, the primary rainbow is visible in the angular range:

$40^\circ $ $\text{to} $ $42^\circ $ $\text{with the horizontal}$

The red colour appears on the outer edge, and the violet colour appears on the inner edge of the primary rainbow.

(iii) Explain the formation of a secondary rainbow. For which angular range with the horizontal is it visible?

Answer:

The secondary rainbow is formed due to two internal reflections of sunlight inside water droplets.

Step-by-step explanation:

Refraction and dispersion:
Sunlight entering the raindrop gets refracted and dispersed into different colours.
Double internal reflection:
The dispersed light undergoes two successive internal reflections inside the raindrop.
Refraction on emergence:
The light refracts again while emerging out of the raindrop.

Due to the extra internal reflection, the order of colours is reversed compared to the primary rainbow.

For the secondary rainbow:

Red light emerges at an angle of approximately $50^\circ$
Violet light emerges at an angle of approximately $53^\circ$

Hence, the secondary rainbow is visible in the angular range:

50^\circ $ $\text{to}$ $53^\circ $ $\text{with the horizontal}

The secondary rainbow is fainter due to loss of light intensity during the second internal reflection.

(iv) Is it possible to see primary and secondary rainbow simultaneously? Under what conditions?

Answer:

Yes, it is possible to see both primary and secondary rainbows simultaneously.

This occurs when:

The raindrops are large and uniformly distributed.
The Sun is bright and at a low altitude.
The observer is positioned such that light undergoing one internal reflection forms the primary rainbow, while light undergoing two internal reflections forms the secondary rainbow.

In this case:

The primary rainbow appears brighter with red on the outer side.
The secondary rainbow appears fainter, outside the primary rainbow, with reversed colour order.

The dark region between the two rainbows is known as Alexander’s dark band.

Question v) (i)

Explain chromatic aberration for spherical lenses. State a method to minimize or eliminate it.

Answer:

Chromatic aberration is a defect of spherical lenses that occurs due to the dispersion of light. The refractive index of a lens material depends on the wavelength (or color) of light. Hence, different colors are refracted by different amounts when passing through the lens.

As a result:

Different colors have different focal lengths.
For a convex lens, violet light focuses closer to the lens, while red light focuses farther.

This causes the image formed by a lens to be blurred and surrounded by colored fringes. This defect is called chromatic aberration.

Let:

$$f_r$$ = focal length for red light
$$f_v$$ = focal length for violet light

Since: $$f_r > f_v$$

The longitudinal chromatic aberration is given by:

$$CA = f_r - f_v$$

Thus, chromatic aberration results in the inability of a lens to focus all colors at a single point.

Method to minimize or eliminate chromatic aberration:

Chromatic aberration can be minimized or eliminated by using an achromatic lens combination, which consists of two lenses made of different materials (e.g., crown glass and flint glass) having different dispersive powers.

Question v) (ii)

What is achromatism? Derive a condition to achieve achromatism for a lens combination. State the conditions for it to be converging.

Answer:

Achromatism is the property of an optical system in which chromatic aberration is eliminated, so that light of different wavelengths (colors) is brought to the same focus.

An optical system that satisfies this condition is called an achromatic combination.

Derivation of condition for achromatism:

Consider two thin lenses in contact:

Lens 1: focal length $$f_1$$ , dispersive power $\omega_1$
Lens 2: focal length $$f_2$$ , dispersive power $\omega_2$

The chromatic aberration produced by a lens is proportional to:

$\dfrac{\omega}{f}$

For achromatism, the total chromatic aberration of the combination must be zero:

$\dfrac{\omega_1}{f_1} + \dfrac{\omega_2}{f_2} = 0$

Rewriting the condition:

$\dfrac{\omega_1}{f_1} = - \dfrac{\omega_2}{f_2}$

This is the condition for achromatism of a lens combination.

Conditions for the achromatic combination to be converging:

One lens must be a converging (convex) lens.
The other lens must be a diverging (concave) lens.
The converging lens should have a smaller dispersive power (e.g., crown glass).
The diverging lens should have a larger dispersive power (e.g., flint glass).
The net focal length of the combination must be positive.

Hence, a converging achromatic combination is obtained by combining a convex lens of crown glass with a concave lens of flint glass such that the above achromatism condition is satisfied.

Question (vi) Describe spherical aberration for spherical lenses. What are different ways to minimize or eliminate it?

Answer:

Definition:

Spherical aberration is an optical defect that occurs in a spherical lens when rays of light passing through different parts of the lens do not converge at the same focal point.

Explanation:

Consider a convex spherical lens. Rays of light parallel to the principal axis fall on the lens:

Rays passing through the central region of the lens are refracted less.
Rays passing through the marginal (outer) region are refracted more.

As a result, marginal rays focus at a point closer to the lens, while paraxial rays focus farther away. Hence, no single sharp focus is formed.

This causes the image to appear blurred instead of sharp.

Diagram Concept (Theory):

Marginal focus lies nearer than the principal focus for a convex lens.

Causes of Spherical Aberration:

Spherical shape of lens surfaces
Large aperture of the lens

Methods to Minimize or Eliminate Spherical Aberration:

Using a stop (aperture):
By blocking marginal rays and allowing only paraxial rays to pass, spherical aberration is reduced.
Using lenses with smaller aperture:
Smaller apertures reduce the difference between marginal and paraxial rays.
Using aspherical lenses:
Aspherical surfaces bring all rays to the same focus.
Using a combination of lenses:
Properly designed lens combinations reduce aberrations.

Conclusion:

Spherical aberration is an unavoidable defect in spherical lenses, but it can be minimized using suitable optical techniques.

Question (vii)

Define and describe magnifying power of an optical instrument. How does it differ from linear or lateral magnification?

Answer:

Definition of Magnifying Power:

Magnifying power of an optical instrument is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the eye when seen directly.

In mathematical form:

$M = \dfrac{\theta_i}{\theta_o}$

where,
$\theta_i$ = angle subtended by the image at the eye
$\theta_o$ = angle subtended by the object at the eye

Explanation:

Magnifying power indicates how much larger an object appears when viewed through an optical instrument such as:

Simple microscope
Compound microscope
Astronomical telescope

It depends on angular size and not on actual size.

Linear or Lateral Magnification:

Linear magnification is defined as the ratio of the height of the image to the height of the object.

$m = \dfrac{h_i}{h_o}$

where,
$$h_i$$ = height of image
$$h_o$$ = height of object

Difference Between Magnifying Power and Linear Magnification:

Magnifying Power	Linear (Lateral) Magnification
Based on angular size	Based on actual size
Used for optical instruments	Used for lenses and mirrors
Related to angles subtended	Related to image and object height
Important for eye observation	Independent of observer’s eye

Conclusion:

Magnifying power describes the apparent enlargement of an object, while linear magnification describes the actual enlargement of an image.

Question (viii) Derive an expression for magnifying power of a simple microscope. Obtain its minimum and maximum values in terms of its focal length.

Answer:

Definition:

A simple microscope is a convex lens of small focal length used to view small objects by forming a magnified virtual image.

Magnifying Power:

Magnifying power ( $M$ ) of a microscope is defined as the ratio of the angle subtended at the eye by the image formed by the microscope to the angle subtended by the object when placed at the least distance of distinct vision ( $D$ ).

M = \frac{\theta_i}{\theta_o}

Case 1: Image formed at Infinity (Maximum Magnifying Power)

For relaxed eye viewing, the final image is formed at infinity.

$$ \theta_i \approx \frac{h}{f} $$

where $h$ is the height of the object and $f$ is the focal length of the lens.

Without microscope, the object at least distance of distinct vision $D$ subtends angle:

$$ \theta_o \approx \frac{h}{D} $$

Therefore, magnifying power:

$$ M_{\max} = \frac{\theta_i}{\theta_o} = \frac{h/f}{h/D} = \frac{D}{f} $$

Maximum magnifying power:

$$ \boxed{M_{\max} = \frac{D}{f}} $$

Case 2: Image formed at Least Distance of Distinct Vision (Minimum Magnifying Power)

For maximum clarity, the image is formed at the least distance of distinct vision $D$ .

Using lens formula:

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

For virtual image at distance $v = -D$ :

$$ \frac{1}{f} = -\frac{1}{D} + \frac{1}{u} $$

Angular magnification is given by:

$$ M = 1 + \frac{D}{f} $$

Minimum magnifying power:

$$ \boxed{M_{\min} = 1 + \frac{D}{f}} $$

Final Results:

Maximum magnifying power: $\displaystyle M_{\max} = \frac{D}{f}$
Minimum magnifying power: $\displaystyle M_{\min} = 1 + \frac{D}{f}$

Exam Tip:

Always mention both cases clearly, write the definition of magnifying power, and box the final expressions to score full marks in board examinations.

ix) Derive the expressions for the magnifying power and the length of a compound microscope using two convex lenses.

Answer:

Compound Microscope:

A compound microscope consists of two coaxially placed convex lenses:

Objective lens: Short focal length $f_o$ , placed near the object.
Eyepiece lens: Larger focal length $f_e$ , acts as a magnifying glass.

The object is placed just beyond the focal point of the objective lens. The objective forms a real, inverted and magnified intermediate image. This image acts as the object for the eyepiece.

Magnifying Power:

Magnifying power ( $M$ ) of a compound microscope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the unaided eye placed at the least distance of distinct vision ( $D$ ).

$M = M_o \times M_e$

Where,

$M_o$ = magnifying power of objective
$M_e$ = magnifying power of eyepiece

Magnification by Objective:

$M_o = \frac{v_o}{u_o}$

For small objects placed near the focal point,

$M_o \approx \frac{L}{f_o}$

where $L$ is the tube length of the microscope.

Magnification by Eyepiece:

For final image at the least distance of distinct vision,

$M_e = 1 + \frac{D}{f_e}$

Total Magnifying Power:

$M = \frac{L}{f_o} \left(1 + \frac{D}{f_e}\right)4

This is the required expression for magnifying power of a compound microscope.

Length (Tube Length) of Compound Microscope:

The length of the microscope is the distance between the objective and eyepiece. It is approximately given by:

$L = v_o + u_e$

Since the intermediate image lies close to the focal point of the eyepiece,

$L \approx f_o + f_e$

Thus, the length of a compound microscope depends on the focal lengths of the objective and eyepiece.

x) What is a terrestrial telescope and an astronomical telescope?

Answer:

Astronomical Telescope:

An astronomical telescope is an optical instrument used to observe distant celestial objects like stars, planets, and galaxies.

It consists of:

A large aperture convex objective lens of long focal length
A convex eyepiece of short focal length

The final image formed is:

Magnified
Inverted
At infinity (for relaxed eye)

Astronomical telescopes are mainly used for scientific and observational purposes.

Terrestrial Telescope:

A terrestrial telescope is used to observe distant objects on the Earth such as buildings, trees, and ships.

It is similar to an astronomical telescope but contains an additional erecting lens or prism system.

The final image formed is:

Magnified
Erect (upright)

Terrestrial telescopes are commonly used for surveying and military observations.

Difference:

Astronomical telescope forms an inverted image.
Terrestrial telescope forms an erect image.

Question (xi) Obtain the expressions for magnifying power and the length of an astronomical telescope under normal adjustments.

Answer:

An astronomical telescope is used to observe distant objects. It consists of two convex lenses:

Objective lens of large focal length $f_o$
Eyepiece lens of small focal length $f_e$

Normal adjustment: The telescope is said to be in normal adjustment when the final image is formed at infinity, so that the eye is relaxed.

Ray Diagram Description:

Parallel rays coming from a distant object fall on the objective lens and form a real, inverted image at its focal plane. This image acts as an object for the eyepiece. The eyepiece is adjusted so that this image lies at the focal plane of the eyepiece, producing parallel rays after refraction.

Magnifying Power:

Magnifying power $M$ is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the unaided eye.

Let,

Angle subtended by the object at the objective, $\alpha \approx \frac{h}{f_o}$

Angle subtended by the final image at the eye, $\beta \approx \frac{h}{f_e}$

Therefore, magnifying power,

$$ M = \frac{\beta}{\alpha} = \frac{h/f_e}{h/f_o} = \frac{f_o}{f_e} $$

The negative sign indicates that the image is inverted.

$$ \boxed{M = -\frac{f_o}{f_e}} $$

Length of the Astronomical Telescope:

In normal adjustment, the distance between the objective and the eyepiece is equal to the sum of their focal lengths.

$$ L = f_o + f_e $$

Hence, the length of the astronomical telescope under normal adjustment is,

$$ \boxed{L = f_o + f_e} $$

Final Results:

Magnifying power: $\( M = -\dfrac{f_o}{f_e} )$
Length of telescope: $$ L = f_o + f_e $$

Obtain the expressions for magnifying power and the length

Question xii) What are the limitations in increasing the magnifying powers of
(i) Simple microscope
(ii) Compound microscope
(iii) Astronomical telescope?

Answer:

The magnifying power of optical instruments cannot be increased indefinitely due to certain practical and physical limitations. These are explained below.

(i) Limitations of magnifying power of a Simple Microscope

The magnifying power $M$ of a simple microscope is given by:

$$ M = 1 + \frac{D}{f} $$

where $D$ is the least distance of distinct vision and $f$ is the focal length of the lens.

Limitations:

To increase magnification, the focal length $f$ must be very small, which is difficult to manufacture.
Very small focal length lenses suffer from spherical and chromatic aberrations.
The image becomes blurred and unclear at high magnification.
The field of view decreases considerably.

(ii) Limitations of magnifying power of a Compound Microscope

The magnifying power of a compound microscope is:

$$ M = \frac{L}{f_o} \times \frac{D}{f_e} $$

where $L$ is the tube length, $f_o$ is focal length of objective and $f_e$ is focal length of eyepiece.

Limitations:

Reducing $f_o$ and $f_e$ increases aberrations.
Diffraction of light limits the resolving power.
Loss of brightness due to absorption and scattering of light.
Very large tube length $L$ is impractical.

(iii) Limitations of magnifying power of an Astronomical Telescope

The magnifying power of an astronomical telescope is:

$$ M = \frac{f_o}{f_e} $$

where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.

Limitations:

Increasing $f_o$ makes the telescope very long and bulky.
Very small $f_e$ causes eye strain and aberrations.
Diffraction limits angular resolution.
Atmospheric disturbances (seeing conditions) reduce image clarity.

Conclusion: The magnifying power of optical instruments is limited by lens aberrations, diffraction of light, loss of brightness, and practical construction difficulties.

✦✦✦ End of Solution ✦✦✦

Key Takeaways:

Always use curly braces {} around numerator and denominator in fractions
\frac{a}{b} is preferred over a \div b for professional mathematical typesetting
Use \boxed{} to emphasize important formulas in your document
For inline fractions in text, consider using / or \nicefrac{}{} for better readability

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Q(i). As per recent understanding, light consists of

Q(ii). Consider the optically denser lenses P, Q, R and S drawn below. According to Cartesian sign convention, which of these have positive focal length?

Q(iii). Two plane mirrors are inclined at an angle of 40° between them. Number of images formed is

Q(iv). A concave mirror of radius of curvature 40 cm produces an image twice the size of the object for shaving. Object distance is

Q(v). Which aberration does NOT occur in spherical mirrors?

Q(vi). A monkey sees a fish at 2.5 m and fish sees monkey at 1.6 m. Actual distances are equal. Refractive index of water is

Q(vii). Total internal reflection is involved in

Q(viii). A person using spectacles of power −2 D uses lenses which are

Q(ix). A spherical marble (μ = 1.5, radius = 1.5 cm) has an air bubble at centre. Its apparent position is

Q(x). Select the WRONG statement.

Q(xi). Prism with maximum dispersive power is

Q(xii). Which is NOT involved in rainbow formation?

Q(xiii). Correct statements for a simple microscope are

Question 2 (i) As per recent development, what is the nature of light? Wave optics and particle nature of light are used to explain which phenomena of light respectively?

Question 2 (ii) Which phenomena can be satisfactorily explained using ray optics? State the assumptions on which ray optics is based.

Question 2 (iii) What is focal power of a spherical mirror or a lens? What may be the reason for using $ \dfrac{1}{f} $ as its expression?

Question (iv)> At which positions of the object do spherical mirrors produce: (i) diminished image, (ii) magnified image?

Question (v) State the restrictions for having images produced by spherical mirrors to be appreciably clear.

Question (vi) Explain spherical aberration for spherical mirrors. How can it be minimized? Can it be eliminated by some curved mirrors?

Question (vii) Define absolute refractive index and relative refractive index. Explain in brief with an illustration for each.

Question (viii) Explain ‘mirage’ as an illustration of refraction.

Question (ix) Under what conditions is total internal reflection possible? Explain it with a suitable example. Define critical angle of incidence and obtain an expression for it.

Question (x) Describe construction and working of an optical fibre. What are the advantages of optical fibre communication over electronic communication?

Question (xi) Why is a prism binoculars preferred over traditional binoculars? Describe its working in brief.

xii) A spherical surface separates two transparent media. Derive an expression that relates object and image distances with the radius of curvature for a point object. Clearly state the assumptions.

Question (xiii) Derive lens makers’ equation. Why is it called so? Under which conditions focal length f and radii of curvature R are numerically equal for a lens?

Question 2 (i) What are different types of dispersion of light? Why do they occur?

Question 2 (ii) Define angular dispersion for a prism. Obtain its expression for a thin prism. Relate it with the refractive indices of the material of the prism for corresponding colours.

Question (iii) Explain and define dispersive power of a transparent material. Obtain its expressions in terms of (i) angles of deviation and (ii) refractive indices.

Question (iv)

(i) State the conditions under which a rainbow can be seen.

(ii) Explain the formation of a primary rainbow. For which angular range with the horizontal is it visible?

(iii) Explain the formation of a secondary rainbow. For which angular range with the horizontal is it visible?

(iv) Is it possible to see primary and secondary rainbow simultaneously? Under what conditions?

Question v) (i)

Question v) (ii)

Question (vi) Describe spherical aberration for spherical lenses. What are different ways to minimize or eliminate it?

Question (vii)

Question (viii) Derive an expression for magnifying power of a simple microscope. Obtain its minimum and maximum values in terms of its focal length.

ix) Derive the expressions for the magnifying power and the length of a compound microscope using two convex lenses.

x) What is a terrestrial telescope and an astronomical telescope?

Question (xi) Obtain the expressions for magnifying power and the length of an astronomical telescope under normal adjustments.

Question xii) What are the limitations in increasing the magnifying powers of (i) Simple microscope (ii) Compound microscope (iii) Astronomical telescope?

(i) Limitations of magnifying power of a Simple Microscope

(ii) Limitations of magnifying power of a Compound Microscope

(iii) Limitations of magnifying power of an Astronomical Telescope

Frequently Asked Questions (FAQ)

Question (iv)> At which positions of the object do spherical mirrors produce:
(i) diminished image,
(ii) magnified image?

Question (xiii) Derive lens makers’ equation. Why is it called so? Under which conditions focal length $f$ and radii of curvature $R$ are numerically equal for a lens?

Question xii) What are the limitations in increasing the magnifying powers of
(i) Simple microscope
(ii) Compound microscope
(iii) Astronomical telescope?

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