Light – Reflection & Refraction

Correct Option: (B) Centre of curvature

The centre of the hollow sphere of glass of which the spherical mirror is a part is called its Centre of Curvature ($C$).

Correct Option: (C) 10 cm

For a spherical mirror of small aperture, $f = \frac{R}{2} = \frac{20cm}{2} = \mathbf{10cm}$.

Correct Option: (C) Concave

A concave mirror produces a virtual, erect, and enlarged image when the object is placed between its pole (P) and principal focus (F). Convex always produces diminished, and plane mirror produces same size.

Correct Option: (C) Convex

A convex mirror always forms a virtual, erect, and diminished image for all positions of the object in front of it.

Correct Option: (A) Both concave

According to Cartesian sign conventions, focal length ($f$) is always negative for both a concave mirror and a concave lens.

Correct Option: (A) Both A and R are true, and R is the correct explanation of A.

Parallel incident rays striking a concave mirror reflect and converge at a single real focal point F on the principal axis in front of the mirror.

Correct Option: (A) Both A and R are true, and R is the correct explanation of A.

Convex mirrors curve outwards, offering a wider field of view and always forming upright (erect) though diminished images of rear traffic.

Answer: The geometric centre of the reflecting surface of a spherical mirror is called its Pole ($P$). It lies on the surface of the mirror.

Answer: The principal focus ($F$) of a concave mirror is a point on its principal axis where all light rays travelling parallel and close to the principal axis actually meet (converge) after reflection from the mirror.

(Diagram features: Concave mirror curved inward, axis line, parallel incoming rays from left reflecting through focus F at distance f from pole P).

Answer: For spherical mirrors of small aperture, the focal length is half of its radius of curvature: $$f = \fracR2 \quad or \quad R = 2f$$

Answer:
1. Type of mirror: Large Concave mirror (converging mirror).
2. Working: The solar furnace is placed at the principal focus of the concave mirror. Parallel incoming sun rays carrying solar thermal energy strike the mirror and converge at the focus, concentrating immense heat and achieving temperatures up to $1800^\circC - 3000^\circC$.

Answer: They share only one common principal axis. The straight line joining their respective poles ($P_1$ and $P_2$) and passing through the common centre of curvature $O$ forms the single shared principal axis.

Answer: The order of surfaces from top to bottom is:

1. Top part: Concave mirror (forms an enlarged erect image of the head when standing close).
2. Middle part: Plane mirror (forms a same-sized image of the middle body).
3. Bottom part: Convex mirror (forms a diminished erect image of the legs).

Answer: Convex mirrors are preferred as rear-view mirrors due to two key reasons:

1. Always Erect Image: They always form an erect (upright), though diminished virtual image of trailing vehicles.
2. Wider Field of View: Because they curve outwards, they provide a much wider field of view compared to plane mirrors, enabling the driver to monitor a larger traffic area.

Answer:
• Centre of curvature ($C$): The centre of the hollow sphere of glass of which the reflecting spherical mirror forms a part.
• Relation: $R = 2f$ or $f = \frac{R}{2}$.
• Numerical example: If a mirror has radius of curvature $R = 30cm$, its focal length is $f = \frac{30}{2} = 15cm$.

Correct Option: (C) Convex

A convex mirror always forms a virtual, erect, and diminished image for all finite and infinite object positions.

Correct Option: (D) Between P and F (Pole and Focus)

When an object is placed between the pole and focus of a concave mirror, rays diverge after reflection and appear to intersect behind the mirror forming an erect, enlarged virtual image.

Correct Option: (B) Real, inverted, same size ($m = -1$) formed at C

When the object is at $C$, the image is also formed at $C$, inverted, real, and with magnification $m = -1$.

Correct Option: (B) Between F and C

When an object is placed between F and C in front of a concave mirror, a real, inverted and enlarged image is formed beyond C ($|m| > 1$).

Correct Option: (A) Both A and R are true, and R is the correct explanation of A.

Since the reflecting surface is curved outwards, parallel rays diverge and appear to meet at the virtual focus behind the mirror.

Answer: Magnification $m = -3$ conveys:

1. Type of mirror: It is a Concave mirror (only concave mirrors form real magnified images).
2. Nature of image: The negative sign indicates the image is Real and Inverted.
3. Size of image: The image is enlarged (3 times the height of the object) ($|m| = 3 > 1$).
4. Position of object: The object lies between the Focus (F) and Centre of curvature (C) of the mirror.

Answer:
(i) Type of mirror: Concave mirror (a convex mirror only gives magnification $0 < m < +1$; a plane mirror gives $m = +1$; only a concave mirror can produce magnified virtual images with $m > +1$).
(ii) Position of object: The object is placed between the Pole (P) and Principal Focus (F) of the concave mirror.
(iii) Characteristics: Image is virtual, erect, enlarged (3 times), and formed behind the mirror.

Answer: Four characteristics of images formed by convex mirrors:

1. Virtual (cannot be obtained on a screen).
2. Erect (upright).
3. Diminished (smaller than object size, $m < +1$).
4. Located behind the mirror (between Pole P and Focus F).

Answer: (1) The image is always virtual and erect. (2) The image is always diminished and formed behind the mirror between P and F.

Answer:
• Ray 1: Ray from tip A parallel to principal axis passes through Focus F after reflection.
• Ray 2: Ray from tip A passing through Centre of Curvature C strikes normally and retraces its path.
• Intersection: Both rays intersect between F and C below the axis to form inverted image $A'B'$.
• Characteristics: Real, Inverted, Diminished, and formed between F and C in front of the mirror.

Answer: A ray of light directed towards the principal focus F of a convex mirror emerges parallel to the principal axis after reflection. The normal at the point of incidence is the radius line passing through C; the angle between incident ray and normal is $\angle i$, and between reflected ray and normal is $\angle r$, where $\angle i = \angle r$.

Answer:
Case (i) Object between P and F:
• Position: Behind the mirror
• Nature & Size: Virtual, Erect, and Enlarged (Magnified).
Case (ii) Object at Centre of Curvature (C):
• Position: At C
• Nature & Size: Real, Inverted, and Same size as object ($m = -1$).

Answer:
• Object Position: The face/teeth must be placed close to the mirror, between the Pole (P) and Principal Focus (F).
• Image Characteristics: Forms an erect, virtual, and magnified (enlarged) image behind the mirror, allowing dentists/shavers to see clear magnified details.

Solutions:
(i) Given: $u = -30cm, f = -15cm, h = +4cm$. $$\frac1v = \frac1f - \frac1u = \frac1{-15} - \frac1{-30} = -\frac115 + \frac130 = -\frac130 \implies \mathbf{v = -30cm}$$
(ii) Since $u = v = -30cm = 2f = R$ (object is at C), the image is Real, Inverted, and Same size ($4cm$) formed at C in front of mirror.
(iii) Ray diagram shows object AB at C, ray 1 parallel through F, ray 2 through F emerging parallel, meeting at C to give inverted image $A'B'$.
(iv) For convex mirror: $f = +15cm, m = +\frac{1}{2} = +0.5$. $$m = -\frac{v}{u} \implies \frac12 = -\frac{v}{u} \implies v = -\fracu2$$ Using mirror formula: $$\frac1{-u/2} + \frac1u = \frac115 \implies -\frac2u + \frac1u = \frac115 \implies -\frac1u = \frac115 \implies \mathbf{u = -15cm}$$ The object is placed at $15cm$ in front of the convex mirror.

Correct Option: (C) +1.0 cm

Diverging mirror (convex) $\implies f = +10cm, u = -30cm$. $$\frac1v = \frac110 - \frac1{-30} = \frac430 \implies v = +7.5cm$$ $$m = -\frac{v}{u} = -\frac{+7.5}{-30} = +0.25 \implies h' = m \cdot h = 0.25 \times 4cm = \mathbf{+1.0cm}$$.

Correct Option: (B) Image is real, inverted and same size as object

Magnification $m = -1$ signifies a negative sign for real & inverted, and magnitude $1$ for identical size ($h' = h$).

Given: $f = -20cm, u = -30cm$.
$$\frac1v + \frac1u = \frac1f \implies \frac1v = \frac1{-20} - \frac1{-30} = -\frac120 + \frac130 = \frac{-3 + 2}60 = -\frac160$$ $$\mathbf{v = -60cm}$$ Since $v$ is negative, the image is formed at $60cm$ in front of the mirror and is Real & Inverted.

Given: $u = -10cm$, real image $\implies m = -3$.
(i) Image distance ($v$): $$m = -\frac{v}{u} \implies -3 = -\fracv{-10} \implies -3 = \fracv10 \implies \mathbf{v = -30cm}$$ The image is formed at $30cm$ in front of the mirror.
(ii) Focal length ($f$): $$\frac1f = \frac1v + \frac1u = \frac1{-30} + \frac1{-10} = \frac{-1 - 3}30 = -\frac430 = -\frac215$$ $$\mathbf{f = -\frac152 = -7.5cm}$$

Answer: Since the image is obtained on a screen, it must be a real and magnified image ($m = -2$). For a concave mirror to produce a real enlarged image, the object must be placed between the Principal Focus (F) and Centre of Curvature (C).

Answers:
(i) Concave mirror (only concave mirrors form real inverted images on a screen with $m = -1$).
(ii) Since image is formed on a screen at $50cm$, $v = -50cm$. With $m = -1$, the object and image are both at C ($v = u = -2f = -50cm$). Therefore, focal length $$f = \fracv2 = \frac{-50}2 = \mathbf{-25cm}$$.
(iii) Object distance $$u = \mathbf{-50cm}$$ (at a distance of 50 cm in front of mirror).

Given: $h = +2cm, u = -30cm$, concave mirror with $R = -40cm \implies f = \frac{R}{2} = -20cm$.
(i) Position of image ($v$): $$\frac1v = \frac1f - \frac1u = \frac1{-20} - \frac1{-30} = -\frac120 + \frac130 = -\frac160 \implies \mathbf{v = -60cm}$$ The image is formed at $60cm$ in front of the concave mirror (beyond C).
(ii) Nature of image: Real and Inverted.
(iii) Size of image ($h'$): $$m = -\frac{v}{u} = -\frac{-60}{-30} = -2 \implies h' = m \cdot h = (-2)(+2cm) = \mathbf{-4cm}$$ The image is enlarged with a height of $4cm$ formed downwards.

New Cartesian Sign Convention:

1. All distances are measured from the Pole (P) as the origin.
2. Distances in the direction of incident light ($+x$) are positive; opposite to incident light ($-x$) are negative.
3. Heights above the principal axis ($+y$) are positive; below ($-y$) are negative.

Answers:
(i) $u = -20cm, f = -10cm \implies \frac{1}{v} = \frac{1}{-10} - \frac{1}{-20} = -\frac{1}{20} \implies \mathbf{v = -20cm}, \mathbf{m = -1}$.
(ii) $u = -25cm, f = -15cm \implies \frac{1}{v} = \frac{1}{-15} - \frac{1}{-25} = -\frac{2}{75} \implies v = -37.5cm$. $m = -\frac{-37.5}{-25} = -1.5 \implies h' = (-1.5)(5cm) = \mathbf{-7.5cm}$ (enlarged inverted).
(iii) $f = +15cm, v = +5cm \implies \frac{1}{u} = \frac{1}{f} - \frac{1}{v} = \frac{1}{15} - \frac{1}{5} = -\frac{2}{15} \implies \mathbf{u = -7.5cm}$.
(iv) $m = -2$ indicates the image is Real, Inverted, and 2 times magnified compared to the object.

Correct Option: (B) Towards the normal ($\angle i > \angle r$)

Light slows down in an optically denser medium, causing the wavefront to pivot towards the interface normal.

Correct Option: (D) It has no unit

Refractive index is a ratio of two identical physical quantities (speeds $n = c/v$), so it is dimensionless.

Correct Option: (C) air > water > glass

$v = \frac{c}{n}$. Lower refractive index corresponds to higher speed of light ($v_{air} \approx 3 \times 10^8\text{ m/s} > v_{water} = 2.25 \times 10^8\text{ m/s} > v_{glass} = 2 \times 10^8\text{ m/s}$).

Correct Option: (C) Along the normal ($\angle i = 0^\circ$)

At normal incidence ($\angle i = 0^\circ$), the angle of refraction is $\angle r = 0^\circ$, meaning lateral displacement is zero.

Correct Option: (A) Both A and R are true, and R is the correct explanation of A.

Change in phase velocity across the interface boundary dictates the direction of refraction according to Snell's law.

Correct Option: (A) Both A and R are true, and R is the correct explanation of A.

According to Snell's law, $\sin r = \frac{\sin 0^\circ}{n} = 0 \implies \angle r = 0^\circ$.

Answer:
• Refraction: The phenomenon of bending of a ray of light when it travels obliquely from one transparent medium to another.
• Cause: Difference in the speed of light in different media.

Laws of Refraction:
1. The incident ray, refracted ray, and the normal at the point of incidence all lie in the same plane.
2. Snell's Law: $\frac{\sin i}{\sin r} = constant = n_{21}$.
Calculation: $$n = \frac{c}{v} \implies v = \fraccn = \frac{3 \times 10^8\text{ m/s}}{1.5} = \mathbf{2 \times 10^8\text{ m/s}}$$.

Answer: The absolute refractive index ($n_m$) of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in that medium: $$n_m = \frac{c}{v} = \frac{\text{Speed of light in vacuum}}{\text{Speed of light in medium}}$$

Given: ${}^y n_x = \frac{n_x}{n_y} = \frac{2}{3}$, ${}^z n_y = \frac{n_y}{n_z} = \frac{4}{3}$, $v_x = 3 \times 10^8\text{ m/s}$.
1. Finding ${}^x n_z$: $${}^z n_x = \frac{n_x}{n_z} = \frac{n_x}{n_y} \times \frac{n_y}{n_z} = \frac23 \times \frac43 = \frac89$$ $${}^x n_z = \frac{1}{{}^z n_x} = \frac{1}{8/9} = \mathbf{\frac98 = 1.125}$$
2. Finding speed of light in medium 'y' ($v_y$): $${}^y n_x = \frac{v_y}{v_x} \implies \frac23 = \frac{v_y}{3 \times 10^8} \implies v_y = \frac23 \times 3 \times 10^8 = \mathbf{2 \times 10^8\text{ m/s}}.$$

Explanation for Glass Slab:
• At the first interface (air to glass), light bends towards the normal ($\angle i > \angle r_1$).
• At the second interface (glass to air), light bends away from the normal by an equal amount ($\angle r_2 < \angle e$).
• Since opposite faces are parallel, the emergent ray is parallel to the incident ray ($\angle i = \angle e$).
• The perpendicular distance between original incident ray path and emergent ray is called lateral displacement.

Answer:
1. Speed: Speed of light increases.
2. Direction: Light ray bends away from the normal ($\angle r > \angle i$).
3. Total Internal Reflection (TIR): Yes, TIR can occur if:
• Light travels from an optically denser to rarer medium.
• Angle of incidence ($i$) is strictly greater than the critical angle ($i > i_c$).

Answer: Light bends the least when the difference in refractive indices ($\Delta n$) between the two media is the smallest.
Comparing differences:
• Between C ($1.43$) and A ($1.55$): $\Delta n = 1.55 - 1.43 = \mathbf{0.12}$ (Smallest!)
• Between D ($1.60$) and B ($1.72$): $\Delta n = 1.72 - 1.60 = 0.12$
Therefore, light will bend the least when passing between Media C and A (or D and B).

Answers:
(i) (1) Incident ray, refracted ray, normal all lie in same plane. (2) Snell's law: $\frac{\sin i}{\sin r} = constant$.
(ii) $v = \frac{c}{n} = \frac{3 \times 10^8}{1.5} = \mathbf{2 \times 10^8\text{ m/s}}$.
(iii) Bends towards the normal because glass ($n = 1.5$) is optically denser than water ($n = 1.33$), causing light to slow down.
(iv) Yes, the emergent ray is strictly parallel to the incident ray because refraction occurs at two parallel opposite interfaces with equal and opposite bending ($\angle i = \angle e$).

Correct Option: (D) Clay

Clay is an opaque material that does not allow light to pass through or undergo refraction.

Correct Option: (C) Optical centre (O)

Near the optical centre, opposite surfaces of a thin lens are parallel, so light passes through without net angular deviation.

Correct Option: (C) A parallel beam of light

Light rays originating from the principal focus of a convex lens emerge parallel to the principal axis after refraction.

Correct Option: (B) At twice the focal length ($2F_1$)

When an object is placed at $2F_1$, a real, inverted image of the same size ($m = -1$) is formed at $2F_2$ on the other side.

Correct Option: (B) Virtual, erect and diminished

A concave lens is a diverging lens and always forms an erect, virtual, and diminished image located between O and $F_1$ on the same side as the object.

Correct Option: (A) Both A and R are true, and R is the correct explanation of A.

A convex lens refracts parallel incident rays towards the central axis, converging them at the real principal focus $F_2$.

Correct Option: (A) Both A and R are true, and R is the correct explanation of A.

A concave lens bends incident rays away from the principal axis, making them appear to diverge from the virtual focus $F_1$.

Answer:
(i) Convex lens (acting as a simple magnifying glass). It is used to produce an enlarged, erect image of palm lines.
(ii) Held close to the palm such that the palm is between the optical centre (O) and principal focus (F₁) of the lens ($u < f$).
(iii) Characteristics: Virtual, Erect, and Enlarged (Magnified), formed on the same side as the palm.

Answer: This happened because the laser light beam was directed precisely through the Optical Centre (O) of the convex lens. The central section of a thin lens has parallel opposite surfaces, so a light ray passing through the optical centre suffers zero deviation and passes straight through.

Summary of 6 Cases for Convex Lens:

1. At Infinity: At $F_2$ • Real & Inverted • Highly diminished (point-sized) ($m \ll -1$).
2. Beyond $2F_1$: Between $F_2$ and $2F_2$ • Real & Inverted • Diminished ($|m| < 1$).
3. At $2F_1$: At $2F_2$ • Real & Inverted • Same size ($m = -1$).
4. Between $F_1$ and $2F_1$: Beyond $2F_2$ • Real & Inverted • Enlarged ($|m| > 1$).
5. At $F_1$: At Infinity • Real & Inverted • Highly enlarged.
6. Between $O$ and $F_1$: Same side as object • Virtual & Erect • Enlarged ($m > +1$).

Summary of 2 Cases for Concave Lens:
1. Object at Infinity: Virtual focus $F_1$ on same side • Virtual & Erect • Highly diminished (point-sized).
2. Object at Finite Distance: Between optical centre O and focus $F_1$ on same side • Virtual & Erect • Diminished ($0 < m < +1$).

Answers:
(i) Virtual, Erect, and Enlarged, formed on the same side as the object.
(ii) $f = +20cm, v = +60cm \implies \frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{60} - \frac{1}{20} = -\frac{2}{60} = -\frac{1}{30} \implies \mathbf{u = -30cm}$ (at 30 cm in front of lens).
(iii) Concave lens $\implies f = -30cm, v = -10cm$. $$\frac1u = \frac1{-10} - \frac1{-30} = -\frac110 + \frac130 = -\frac230 = -\frac115 \implies \mathbf{u = -15cm}$$
(iv) Convex lens, when the object is held close between the optical centre O and focus F ($u < f$).

Correct Options: Both (C) and (D) produce magnified images

Between F and 2F yields a real enlarged image; between O and F yields a virtual magnified image.

Correct Option: (D) −40 cm

Screen $\implies$ real inverted image $\implies m = -3, f = +30cm$. $$m = \frac{v}{u} = -3 \implies v = -3u$$ $$\frac1v - \frac1u = \frac1f \implies \frac1{-3u} - \frac1u = \frac130 \implies \frac{-4}3u = \frac130 \implies 3u = -120 \implies \mathbf{u = -40cm}$$.

Correct Option: (C) $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Correct Option: (A) Both A and R are true, and R is the correct explanation of A.

For virtual erect images, both $u$ and $v$ lie in the negative direction (or $h'$ and $h$ are positive), rendering magnification positive ($m > 0$).

Given: $h = +2mm = +0.2cm, u = -5cm, f = +10cm$.
(i) Image distance ($v$): $$\frac1v = \frac1f + \frac1u = \frac110 + \frac1{-5} = \frac{1 - 2}10 = -\frac110 \implies \mathbf{v = -10cm}$$
(ii) Size of image ($h'$): $$m = \frac{v}{u} = \frac{-10}{-5} = +2 \implies h' = m \cdot h = (+2)(2mm) = \mathbf{+4mm = +0.4cm}$$
(iii) Characteristics: Virtual, Erect, and 2 times magnified, formed on the same side at 10 cm from lens.

Given: Concave lens $\implies f = -15cm, v = -10cm$. $$\frac1u = \frac1v - \frac1f = \frac1{-10} - \frac1{-15} = -\frac110 + \frac115 = -\frac130 \implies \mathbf{u = -30cm}$$ $$m = \frac{v}{u} = \frac{-10}{-30} = \mathbf{+\frac13 \approx +0.33}$$ Object should be placed at $30cm$ in front of the lens. Magnification is $+0.33$.

Given: $h = +5cm, u = -20cm, R = +30cm \implies f = +15cm$. $$\frac1v = \frac1f - \frac1u = \frac115 - \frac1{-20} = \frac115 + \frac120 = \frac760 \implies \mathbf{v = +\frac607 \approx +8.57cm}$$ $$m = -\frac{v}{u} = -\frac{+60/7}{-20} = +\frac37 \approx +0.43$$ $$h' = m \cdot h = \frac37 \times 5cm = \mathbf{+\frac157 \approx +2.14cm}$$ Image is Virtual & Erect, formed at $8.57cm$ behind the mirror with height $2.14cm$.

Given: $h = +5cm, u = -25cm, f = +10cm$. $$\frac1v = \frac1f + \frac1u = \frac110 + \frac1{-25} = \frac{5 - 2}50 = \frac350 \implies \mathbf{v = +\frac503 \approx +16.67cm}$$ $$m = \frac{v}{u} = \frac{+50/3}{-25} = -\frac23 \approx -0.67$$ $$h' = m \cdot h = \left(-\frac23\right)(5cm) = \mathbf{-3.33cm}$$ Image is Real & Inverted, formed at $16.67cm$ on the other side, diminished to $3.33cm$.

Answer:
1. Complete Image: Yes, the lens will still form a complete image of the object because light rays from every point of the object pass through the uncovered half of the lens and refract to intersect at the focal plane.
2. Brightness / Intensity: The brightness of the image will be reduced to half because the number of light rays passing through the lens to form the image is halved.

Calculation:
1. Power of individual lenses: $$P_1 = \frac1{+0.25m} = +4D, \quad P_2 = \frac1{-0.10m} = -10D$$ $$P_net = P_1 + P_2 = +4 - 10 = \mathbf{-6D}$$ Equivalent focal length $$F = \frac1{-6D} = -\frac16m = -\frac1006cm = -\frac503cm \approx -16.67cm$$.
2. Finding image position for $u = -40cm$: $$\frac1v = \frac1F + \frac1u = -\frac350 + \frac1{-40} = -\frac350 - \frac140 = \frac{-12 - 5}200 = -\frac17200$$ $$\mathbf{v = -\frac20017 \approx -11.76cm}$$ Image is formed at $11.76cm$ in front of the lens combination and is virtual & erect.

(a) Rules for Lenses:
1. Ray parallel to principal axis passes through $F_2$ (convex) or appears to diverge from $F_1$ (concave).
2. Ray through $F_1$ (convex) or directed towards $F_2$ (concave) emerges parallel to principal axis.
3. Ray through Optical Centre O passes undeviated.
(b) Numerical Calculation: $h = +4cm, u = -30cm, f = +20cm$. $$\frac1v = \frac120 + \frac1{-30} = \frac{3 - 2}60 = \frac160 \implies \mathbf{v = +60cm}$$ $$m = \frac{v}{u} = \frac{+60}{-30} = -2 \implies h' = (-2)(+4cm) = \mathbf{-8cm}$$ Image is formed at $60cm$ on other side of lens; it is Real, Inverted, and 2 times enlarged ($8cm$).

Answers:
(i) $f = +20cm, v = +60cm \implies \frac{1}{u} = \frac{1}{60} - \frac{1}{20} = -\frac{2}{60} = -\frac{1}{30} \implies \mathbf{u = -30cm}, \mathbf{m = -2}$.
(ii) Concave lens $\implies f = -30cm, u = -15cm$. $$\frac1v = \frac1{-30} + \frac1{-15} = -\frac330 = -\frac110 \implies \mathbf{v = -10cm}$$ $$m = \frac{-10}{-15} = +\frac23 \implies h' = \left(+\frac23\right)(3cm) = \mathbf{+2cm}$$ (virtual erect diminished).
(iii) $m = +0.5$ means image is Virtual, Erect, and half the size ($0.5imes$) of the object.
(iv) Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. Distances measured from optical centre O; $u$ is always negative; $f$ is positive for convex and negative for concave.

Correct Option: (B) +4 D

$f = +25cm = +0.25m \implies P = \frac{1}{+0.25} = \mathbf{+4D}$.

Correct Option: (B) Dioptre (D)

$1ext{ Dioptre (D)} = 1m^{-1}$, defined as the power of a lens of focal length 1 metre.

Correct Option: (B) Concave (diverging lens)

A negative focal length ($f = 1/P = -0.25m = -25cm$) corresponds to a diverging (concave) lens.

Correct Option: (C) $P_{combined} = P_P + P_Q$

The net optical power of thin lenses placed in contact is the algebraic sum of their individual powers.

Correct Option: (A) Both A and R are true, and R is the correct explanation of A.

By sign convention, $f > 0$ for convex lenses, hence $P = 1/f > 0$.

Correct Option: (A) Both A and R are true, and R is the correct explanation of A.

Linear combination of reciprocal focal lengths directly yields $P_{net} = P_1 + P_2$.

Answer:
• Definition: The power of a lens is the measure of its ability to converge or diverge light rays falling on it. It is defined as the reciprocal of its focal length expressed in metres.
• Formula: $$P = \frac{1}{f\text{ (in metres)}} \quad \text{or} \quad P = \frac{100}{f\text{ (in cm)}}$$
• SI Unit: Dioptre (D) ($1D = 1m^{-1}$).

Given: $P = +1.5D$. $$f = \frac1P = \frac1{+1.5D} = +\frac1015m = \mathbf{+\frac23m \approx +0.67m = +66.7cm}$$ • Nature: Converging (Convex) lens. • Corrects: Hypermetropia (Farsightedness).

Answer:
• Student 1: $f_1 = +50cm = +0.5m \implies P_1 = \frac{1}{+0.5} = \mathbf{+2.0D}$ (Converging / Convex lens).
• Student 2: $f_2 = -50cm = -0.5m \implies P_2 = \frac{1}{-0.5} = \mathbf{-2.0D}$ (Diverging / Concave lens).

Calculation:
• Net Power: $$P_net = P_1 + P_2 = (+2D) + (-1D) = \mathbf{+1.0D}$$
• Equivalent Focal Length: $$F = \frac1{P_net} = \frac1{+1.0D} = \mathbf{+1.0m = +100cm}$$
• Since the net power is positive, the combination behaves as a Convex (Converging) lens.

Answers:
(i) $f = -25cm = -0.25m \implies P = \frac{1}{-0.25} = \mathbf{-4.0D}$ (Concave / Diverging lens).
(ii) $P_{net} = +3 - 1 = +2D \implies F = \frac{1}{+2} = \mathbf{+0.5m = +50cm}$.
(iii) Since power is positive (+), the lens is convex, used to correct Hypermetropia (Long-sightedness).
(iv) $f_1 = 0.4m \implies P_1 = 2.5D$; $f_2 = 0.2m \implies P_2 = 5.0D$; $$P_net = 2.5 + 5.0 = \mathbf{+7.5D}$$.

Answer Key:
(a) Laws of reflection: (1) $\angle i = \angle r$. (2) Incident ray, normal, and reflected ray all lie in the same plane.
(b) Numerical: $u = -10cm, f = -15cm$. $$\frac1v = \frac1{-15} - \frac1{-10} = -\frac115 + \frac110 = \frac130 \implies \mathbf{v = +30cm}$$ $$m = -\frac{v}{u} = -\frac{+30}{-10} = \mathbf{+3}$$
(c) Characteristics: The image is formed at $30cm$ behind the mirror; it is Virtual, Erect, and 3 times enlarged.

Answer Key:
(a) Definitions: (i) Pole P = geometric centre of mirror surface; (ii) Centre of curvature C = centre of sphere; (iii) Focus F = point where parallel rays meet; (iv) Focal length f = distance from P to F.
(b) Numerical: $h = +6cm, u = -30cm, R = -20cm \implies f = -10cm$. $$\frac1v = \frac1{-10} - \frac1{-30} = -\frac110 + \frac130 = -\frac230 = -\frac115 \implies \mathbf{v = -15cm}$$ $$m = -\frac{-15}{-30} = -0.5 \implies h' = (-0.5)(+6cm) = \mathbf{-3cm}$$ Image is Real, Inverted, Diminished ($3cm$), formed at $15cm$ in front of mirror (between F and C).

Answer Key:
(b) Calculations: Given $n_w = \frac{4}{3}, n_g = \frac{3}{2}, v_g = 2 \times 10^8\text{ m/s}$.
1. Speed of light in vacuum ($c$): $$c = n_g \times v_g = \frac32 \times 2 \times 10^8 = \mathbf{3 \times 10^8\text{ m/s}}$$
2. Speed of light in water ($v_w$): $$v_w = \frac{c}{n_w} = \frac{3 \times 10^8}{4/3} = \frac94 \times 10^8 = \mathbf{2.25 \times 10^8\text{ m/s}}$$
3. Refractive index of water with respect to glass (${}^g n_w$): $${}^g n_w = \frac{n_w}{n_g} = \frac{4/3}{3/2} = \mathbf{\frac89 \approx 0.89}$$

Answer Key:
(b) Numerical: $h = +3cm, u = -14cm, f = +10cm$. $$\frac1v = \frac110 + \frac1{-14} = \frac110 - \frac114 = \frac{7 - 5}70 = \frac270 = \frac135 \implies \mathbf{v = +35cm}$$ $$m = \frac{v}{u} = \frac{+35}{-14} = -2.5 \implies h' = (-2.5)(+3cm) = \mathbf{-7.5cm}$$ Image is formed at $35cm$ on the other side (beyond 2F); it is Real, Inverted, and Enlarged ($7.5cm$).

Answer Key:
(a) $f = -20cm, h = +4cm, u = -30cm$. $$\frac1v = \frac1{-20} - \frac1{-30} = -\frac160 \implies \mathbf{v = -60cm}$$ $$m = -\frac{-60}{-30} = \mathbf{-2}$$ $$h' = m \cdot h = (-2)(+4cm) = \mathbf{-8cm}$$
(b) Four Characteristics: (1) Real, (2) Inverted, (3) Enlarged (2 times, height 8 cm), (4) Formed at 60 cm in front of mirror (beyond C).

Answer Key for (c):
Convex mirror $\implies f = +15cm, u = -30cm, h = +5cm$. $$\frac1v = \frac115 - \frac1{-30} = \frac115 + \frac130 = \frac330 = \frac110 \implies \mathbf{v = +10cm}$$ $$m = -\frac{+10}{-30} = \mathbf{+\frac13 \approx +0.33}$$ $$h' = m \cdot h = \frac13 \times 5cm = \mathbf{+1.67cm}$$ Image is formed at $10cm$ behind the mirror; it is Virtual, Erect, and Diminished ($1.67cm$).

Answer Key:
(b) $f_1 = +0.20m \implies P_1 = +5D$; $f_2 = -0.30m \implies P_2 = -\frac{10}{3}D \approx -3.33D$.
(i) Combined power $$P_net = 5 - \frac103 = \mathbf{+\frac53D \approx +1.67D}$$
(ii) Equivalent focal length $$F = \frac1{P_net} = \frac35m = \mathbf{+60cm}$$
(iii) Nature: Convex (Converging) lens.
(c) For $u = -40cm, F = +60cm$: $$\frac1v = \frac160 + \frac1{-40} = \frac{2 - 3}120 = -\frac1120 \implies \mathbf{v = -120cm}$$ Image is virtual and erect, formed at $120cm$ in front of the combination.

Answer Key:
(a) When object is beyond C, the image is formed between F and C (Real, Inverted & Diminished).
(c) Numerical: $u = -40cm, f = -15cm$. $$\frac1v = \frac1{-15} - \frac1{-40} = -\frac115 + \frac140 = \frac{-8 + 3}120 = -\frac5120 = -\frac124 \implies \mathbf{v = -24cm}$$ Image is formed at $24cm$ in front of the concave mirror (between $F=15$ and $C=30$). Nature is Real, Inverted, and Diminished.

Answer Key:
(a) Comparison:
• Mirror: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, $m = -\frac{v}{u}$.
• Lens: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, $m = +\frac{v}{u}$.
(b) Concave lens: $u = -40cm, f = -20cm$. $$\frac1v = \frac1{-20} + \frac1{-40} = -\frac340 \implies \mathbf{v = -\frac403 \approx -13.33cm}$$ $$m = \frac{-40/3}{-40} = \mathbf{+\frac13 \approx +0.33}$$ (Virtual, Erect & Diminished).
(c) Convex lens: $u = -40cm, f = +20cm \implies u = -2f$ (object at $2F_1$). $$\frac1v = \frac120 + \frac1{-40} = \frac140 \implies \mathbf{v = +40cm}$$ $$m = \frac{+40}{-40} = \mathbf{-1}$$ (Real, Inverted, and Same size formed at $2F_2$).

Answer Key:
(a) Definitions of Refraction, Refractive Index $n = c/v$, and Snell's Law $\frac{\sin i}{\sin r} = n_{21}$.
(b) $\sin r = \frac{\sin 30^\circ}{1.5} = \frac{0.5}{1.5} = \frac{1}{3} \approx 0.333 \implies \mathbf{r = \arcsin(0.333) \approx 19.47^\circ}$.
(c) $v_{diamond} = \frac{c}{n} = \frac{3 \times 10^8}{2.42} \approx \mathbf{1.24 \times 10^8\text{ m/s}}$.
(d) Light rays originating from the bottom of the pool travel from water (denser) to air (rarer) and bend away from the normal. When these diverging rays enter the observer's eyes, their backward projections meet at a higher point, creating a raised virtual image and causing the pool to appear shallower (apparent depth < real depth).