CG Board Class 10 Science Chapter 13 Solutions: Light: Reflection and Refraction at Spherical Surfaces (प्रकाश: परावर्तन एवं अपवर्तन गोलीय सतह से) (2026-27)

(i) Concave lens is:**

• (a) Only diverging
• (b) Only converging
• (c) Neither converging nor diverging
• (d) Both converging & diverging

(a) Only diverging

Explanation: A concave lens diverges all incident light rays parallel to its principal axis.

Answer: (a) Concave

Explanation: Only a concave mirror forms real-inverted images for distances beyond focus and a magnified erect virtual image when the object is between pole and focus.

Answer: (a) Smaller than object

Explanation: Convex mirrors always form virtual, erect, and diminished (smaller) images.

Answer: (a) Smaller and virtual / (Official textbook key option (a))

Answer: (a) Infinity / (Official textbook key option (a))

Answer: (c) More than 3 cm

Explanation: A convex lens acts as a magnifying glass and forms an erect, magnified image when the object distance (u = 3 cm) is less than its focal length (f). Therefore, f > 3 cm.

(i) The image formed by a convex mirror is always …………………… and ……………………, in all cases.**

(ii) To obtain a real image of the same size as the object from a convex lens, the object must be placed at ……………………

(iii) The power of a lens is +5.0 D. The focal length of the lens will be …………………… cm.

(iv) The focal length of a convex lens is 25 cm. The power of this lens will be …………………… D.

• (i) Small (diminished) and Virtual
• (ii) Centre of curvature (2F₁)
• (iii) 20 cm [Calculation: f = 100/P = (100)/(+5.0) = 20 cm]
• (iv) +4 D [Calculation: P = 100/f = 100/25 = +4 D]

For a spherical mirror with small aperture, the focal length (f) is equal to half of its radius of curvature (R)

f = R/2 or R = 2f

1. Magnification less than 1 (m < 1):

• Convex Mirror: Always forms diminished images (m < 1).
• Concave Mirror: When object is placed beyond C.

1. Magnification equal to 1 (m = 1):

• Plane Mirror: Always forms same-sized images (m = 1).
• Concave Mirror: When object is placed at C.

1. Magnification greater than 1 (m > 1):

• Concave Mirror: When object is placed between C and F, or between F and Pole P.

Convex mirrors are used as rear-view mirrors in vehicles because

1. They always form an erect (upright) image of trailing traffic.
2. Being curved outwards, they offer a much wider field of view compared to plane or concave mirrors, allowing drivers to view a large area behind them.

A Concave Mirror must be used, because only a concave mirror can form real images that can be focused and projected onto a screen.

Concave Mirror is Converging:** Parallel rays reflected from its inner curved surface intersect at a single real focal point (F).

• Convex Mirror is Diverging: Parallel rays reflected from its outer curved surface spread apart (diverge) as if coming from a virtual focal point (F) behind the mirror.

(i) Centre of curvature (ii) Radius of curvature (iii) Pole (iv) Aperture**

(i) (ii) (iii) (iv)

• (i) Centre of Curvature (C): The center of the hollow glass sphere of which the spherical mirror forms a part.
• (ii) Radius of Curvature (R): The radius of the hollow glass sphere of which the mirror forms a part (distance PC).
• (iii) Pole (P): The geometric center point of the reflecting surface of the spherical mirror.
• (iv) Aperture: The effective diameter or linear boundary width of the reflecting surface of the mirror.
• (i) (C): ,
• (ii) (R): ,
• (iii) (P):
• (iv) (Aperture):

Converging Nature (Convex Lens): A convex lens is thicker in the middle than at the edges. When parallel rays pass through it, refraction bends rays towards the principal axis, focusing them at a real focus point (F₂). Hence, a convex lens is a converging lens**.

• Diverging Nature (Concave Lens): A concave lens is thinner in the middle than at the edges. When parallel rays pass through it, refraction bends rays away from the principal axis, spreading them apart as if originating from a virtual focus point (F₁). Hence, a concave lens is a diverging lens.

The Power of a lens (P) is defined as the measure of its ability to converge or diverge light rays falling on it. Mathematically, it is the reciprocal of the focal length (f) measured in meters

P = (1)/(f (in meters))

• SI Unit: Diopter (D) (1 D = 1 m^-1).

Cartesian Sign Convention for Lenses

1. The optical centre (O) of the lens is taken as the origin point.
2. Light rays are always assumed to travel from left to right.
3. Distances measured in the direction of incident light (+X axis, right of O) are positive.
4. Distances measured opposite to the direction of incident light (-X axis, left of O) are negative.
5. Heights measured perpendicularly above the principal axis (+Y axis) are positive.
6. Heights measured perpendicularly below the principal axis (-Y axis) are negative.

• Focal length of convex lens is always positive, and concave lens is always negative.

Case 1: Convex Lens (f = +50 cm = +0.5 m):**

P = (1)/(f (m)) = (1)/(+0.5) = +2.0 D

• Case 2: Concave Lens (f = -50 cm = -0.5 m):

P = (1)/(f (m)) = (1)/(-0.5) = -2.0 D

Given:** Concave mirror: f = -10 cm, u = -15 cm.

• Mirror Formula:

1/f = 1/v + 1/u ⇒ 1/v = 1/f – 1/u

1/v = 1 – 10 – 1 – 15 = -1/10 + 1/15 = (-3 + 2)/(30) = -1/30

v = -30 cm

• Magnification:

m = -v/u = –30 – 15 = -2

• Result: The image is formed at a distance of 30 cm in front of the mirror, is real and inverted, and is 2 times magnified (twice the size of object).

Given:** Convex mirror: R = +30 cm ⇒ f = +30/2 = +15 cm, u = -10 cm, h = 5 cm.

• Mirror Formula:

1/v = 1/f – 1/u = 1/15 – 1 – 10 = 1/15 + 1/10 = (2 + 3)/(30) = 5/30 = 1/6

v = +6 cm

• Height of Image (h’):

m = h’/h = -v/u ⇒ h’ = 5 × (-6 – 10) = 5 × 0.6 = +3 cm

m = +3/5 = +0.6

• Result: Image is formed 6 cm behind the mirror, is virtual and erect, with height 3 cm and magnification +0.6.

Given:** Concave mirror f = -10 cm, magnitude of magnification |m| = 5.

• Case (i): Real Image (m = -5):

m = -v/u = -5 ⇒ v = 5u

1/f = 1/v + 1/u ⇒ 1 – 10 = 1/5u + 1/u = 6/5u

-5u = 60 ⇒ u = -12 cm

• Case (ii): Virtual Image (m = +5):

m = -v/u = +5 ⇒ v = -5u

1 – 10 = 1 – 5u + 1/u = (-1 + 5)/(5u) = 4/5u

-5u = 40 ⇒ u = -8 cm

Part 1: Convex Mirror (R = +30 cm ⇒ f = +15 cm, u = -12 cm):**

1/v = 1/15 – 1 – 12 = 1/15 + 1/12 = (4 + 5)/(60) = 9/60

v = 60/9 = +6.67 cm

Image is virtual, erect, diminished (m = +0.55), formed 6.67 cm behind mirror.

• Part 2: Concave Mirror (R = -30 cm ⇒ f = -15 cm, u = -12 cm):

1/v = 1 – 15 – 1 – 12 = -1/15 + 1/12 = (-4 + 5)/(60) = 1/60

v = +60 cm

Image is virtual, erect, magnified (m = +5), formed 60 cm behind mirror.

Given:** Convex mirror: u = -30 cm, v = +10 cm (behind mirror).

• Mirror Formula:

1/f = 1/v + 1/u = 1/10 + 1 – 30 = (3 – 1)/(30) = 2/30 = 1/15

f = +15 cm

Given:** Concave mirror f = -12 cm, object at focus u = -12 cm.

• Calculation:

1/v = 1/f – 1/u = 1 – 12 – 1 – 12 = -1/12 + 1/12 = 0 ⇒ v = ∞ (Infinity)

• Result: The image will be formed at infinity (∞), as a highly magnified, real, and inverted beam of parallel rays.

Given:** Convex lens f = +15 cm, real image magnification m = -3.

• Lens Magnification:

m = v/u = -3 ⇒ v = -3u

• Lens Formula:

1/f = 1/v – 1/u ⇒ 1/15 = 1 – 3u – 1/u = (-1 – 3)/(3u) = -4/3u

3u = -60 ⇒ u = -20 cm

Given:** Concave lens: f = -30 cm, u = -30 cm, h = 30 cm.

• Lens Formula:

1/v = 1/f + 1/u = 1 – 30 + 1 – 30 = -2/30 = -1/15

v = -15 cm

• Size of Image (h’):

m = h’/h = v/u ⇒ h’ = 30 × (-15 – 30) = +15 cm

• Result: Image is formed at 15 cm on the same side as object (between focus and optical center), is virtual and erect, with height 15 cm.

Given:** Concave lens u = -30 cm, magnification m = +2/3 (virtual image).

• Lens Magnification:

m = v/u = 2/3 ⇒ v = 2/3 × (-30) = -20 cm

• Lens Formula:

1/f = 1/v – 1/u = 1 – 20 – 1 – 30 = -1/20 + 1/30 = (-3 + 2)/(60) = -1/60

f = -60 cm

Given:** Convex lens f = +50 cm.

• Case (i): u = -25 cm:

1/v = 1/f + 1/u = 1/50 + 1 – 25 = (1 – 2)/(50) = -1/50

v = -50 cm (Virtual image on the same side at 50 cm)

• Case (ii): u = -75 cm:

1/v = 1/50 + 1 – 75 = (3 – 2)/(150) = 1/150

v = +150 cm (Real image on the opposite side at 150 cm)

Given:** Power P = +1.5 D.

• Calculation:

f = 1/P = (1)/(+1.5) m = 100/1.5 cm = +66.67 cm (or ≈ 50 cm as per textbook key note)

(Note: Mathematically f = +66.67 cm; lens is convex).

Given:** Concave lens f = -20 cm = -0.2 m.

• Calculation:

P = (1)/(f (m)) = (1)/(-0.2 m) = -5.0 D