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Sign Convention Reminder

• Light is assumed to travel from left to right.
• Object distance \(u\) → always negative.
• Convex lens focal length \(f\) → positive.
• Concave lens focal length \(f\) → negative.
• Real image distance \(v\) → positive.
• Virtual image distance \(v\) → negative.

Lens formula: \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] Magnification: \[ m=\frac{v}{u} \]

1. An object is placed 40 cm in front of a convex lens of focal length 20 cm. Find image distance.
Hint: Convex lens → \(f\) positive. Object in front → \(u\) negative.

Show Solution

\[ f=+20\,cm \] \[ u=-40\,cm \] \[ \frac{1}{20}=\frac{1}{v}-\frac{1}{(-40)} \] \[ \frac{1}{20}=\frac{1}{v}+\frac{1}{40} \] \[ \frac{1}{v}=\frac{1}{20}-\frac{1}{40} \] \[ \frac{1}{v}=\frac{1}{40} \] \[ v=40\,cm \] Image is real and inverted.

2. An object is placed 12 cm in front of a convex lens of focal length 18 cm. Find image distance.
Hint: Object inside focus → image virtual.

Show Solution

\[ f=18 \] \[ u=-12 \] \[ \frac{1}{18}=\frac{1}{v}-\frac{1}{(-12)} \] \[ \frac{1}{18}=\frac{1}{v}+\frac{1}{12} \] \[ \frac{1}{v}=\frac{1}{18}-\frac{1}{12} \] \[ \frac{1}{v}=-\frac{1}{36} \] \[ v=-36\,cm \] Image is virtual and erect.

3. A concave lens has focal length 10 cm. An object is placed 15 cm in front of it. Find image distance.
Hint: Concave lens → focal length negative.

Show Solution

\[ f=-10 \] \[ u=-15 \] \[ \frac{1}{-10}=\frac{1}{v}-\frac{1}{(-15)} \] \[ \frac{1}{-10}=\frac{1}{v}+\frac{1}{15} \] \[ \frac{1}{v}=-\frac{1}{10}-\frac{1}{15} \] \[ \frac{1}{v}=-\frac{5}{30} \] \[ v=-6\,cm \] Image is virtual and diminished.

4. An image is formed at 30 cm by a convex lens of focal length 10 cm. Find object distance.
Hint: Real image → \(v\) positive.

Show Solution

\[ v=+30 \] \[ f=+10 \] \[ \frac{1}{10}=\frac{1}{30}-\frac{1}{u} \] \[ \frac{1}{u}=\frac{1}{30}-\frac{1}{10} \] \[ \frac{1}{u}=-\frac{2}{30} \] \[ u=-15\,cm \]

5. A convex lens produces an image twice the size of the object. Object distance = 20 cm. Find image distance.
Hint: Use magnification formula.

Show Solution

\[ m=2 \] \[ m=\frac{v}{u} \] \[ 2=\frac{v}{-20} \] \[ v=-40\,cm \] Image is virtual.

6. Object height = 3 cm. Magnification = −2. Find image height.

Show Solution

\[ m=\frac{h_i}{h_o} \] \[ -2=\frac{h_i}{3} \] \[ h_i=-6\,cm \] Image is inverted.

7. A convex lens forms a real image at 24 cm when object is at 12 cm. Find focal length.

Show Solution

\[ u=-12 \] \[ v=24 \] \[ \frac{1}{f}=\frac{1}{24}-\frac{1}{-12} \] \[ \frac{1}{f}=\frac{1}{24}+\frac{1}{12} \] \[ \frac{1}{f}=\frac{3}{24} \] \[ f=8\,cm \]

8. A concave lens produces an image at 12 cm when object is at 24 cm. Find focal length.

Show Solution

\[ u=-24 \] \[ v=-12 \] \[ \frac{1}{f}=\frac{1}{-12}-\frac{1}{-24} \] \[ \frac{1}{f}=-\frac{1}{24} \] \[ f=-24\,cm \]

9. A convex lens produces an image at infinity. Where is the object placed?

Show Solution

Object is placed at the **principal focus (F)**.

10. If object is placed at infinity in front of a convex lens, where will image form?

Show Solution

Image forms at the **principal focus of the lens**.

Important Sign Convention for Lenses

• Light is assumed to travel from left to right.
• Distances measured opposite to incident light → negative.
• Distances measured along incident light → positive.
• Object distance \(u\) → always negative.
• Convex lens focal length \(f\) → positive.
• Concave lens focal length \(f\) → negative.
• Real image distance \(v\) → positive.
• Virtual image distance \(v\) → negative.

Question 1

An object is placed 30 cm in front of a convex lens of focal length 20 cm. Find the image distance and nature of the image. Hint: Convex lens → \(f\) positive. Object in front → \(u\) negative.

Solution

\[ f = +20\,cm \] \[ u = -30\,cm \] Lens formula \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] \[ \frac{1}{20}=\frac{1}{v}-\frac{1}{(-30)} \] \[ \frac{1}{20}=\frac{1}{v}+\frac{1}{30} \] \[ \frac{1}{v}=\frac{1}{20}-\frac{1}{30} \] LCM = 60 \[ \frac{1}{v}=\frac{3-2}{60} \] \[ v=60\,cm \] Image is real and inverted.

Question 2

An object is placed 10 cm in front of a convex lens of focal length 15 cm. Find the image distance. Hint: Object is inside focus → image becomes virtual.

Solution

\[ f=+15 \] \[ u=-10 \] \[ \frac{1}{15}=\frac{1}{v}-\frac{1}{(-10)} \] \[ \frac{1}{15}=\frac{1}{v}+\frac{1}{10} \] \[ \frac{1}{v}=\frac{1}{15}-\frac{1}{10} \] \[ \frac{1}{v}=\frac{2-3}{30} \] \[ v=-30\,cm \] Negative \(v\) means image is virtual.

Question 3

A concave lens has focal length 15 cm. An object is placed 20 cm in front of the lens. Find image distance. Hint: Concave lens → focal length negative.

Solution

\[ f=-15 \] \[ u=-20 \] \[ \frac{1}{-15}=\frac{1}{v}-\frac{1}{(-20)} \] \[ \frac{1}{-15}=\frac{1}{v}+\frac{1}{20} \] \[ \frac{1}{v}=-\frac{1}{15}-\frac{1}{20} \] \[ \frac{1}{v}=-\frac{7}{60} \] \[ v=-8.6\,cm \] Image is virtual and erect.

Question 4

Find the magnification when image distance = 40 cm and object distance = 20 cm.

Solution

\[ m=\frac{v}{u} \] \[ m=\frac{40}{-20} \] \[ m=-2 \] Image is inverted.

Question 5

A convex lens produces an image at 30 cm when the object is placed 15 cm away. Find focal length.

Solution

\[ u=-15 \] \[ v=+30 \] \[ \frac{1}{f}=\frac{1}{30}-\frac{1}{-15} \] \[ \frac{1}{f}=\frac{1}{30}+\frac{1}{15} \] \[ \frac{1}{f}=\frac{3}{30} \] \[ f=10\,cm \]

Question 6

An object is placed at 25 cm from a convex lens of focal length 10 cm. Find image distance.

Solution

\[ f=10 \] \[ u=-25 \] \[ \frac{1}{10}=\frac{1}{v}+\frac{1}{25} \] \[ \frac{1}{v}=\frac{1}{10}-\frac{1}{25} \] \[ \frac{1}{v}=\frac{3}{50} \] \[ v=16.7\,cm \]

Question 7

A concave lens of focal length 20 cm forms an image at 10 cm. Find object distance.

Solution

\[ f=-20 \] \[ v=-10 \] \[ \frac{1}{-20}=\frac{1}{-10}-\frac{1}{u} \] \[ \frac{1}{u}=-\frac{1}{20} \] \[ u=-20\,cm \]

Question 8

Object height = 2 cm Magnification = 3 Find image height.

Solution

\[ m=\frac{h_i}{h_o} \] \[ 3=\frac{h_i}{2} \] \[ h_i=6\,cm \]

Question 9

Object is placed at 2f from convex lens. Describe image.

Answer

Image forms at 2f Real Inverted Same size

Question 10

Why does a concave lens always produce virtual image?

Answer

Because the refracted rays diverge and appear to meet behind the lens. Hence the image is always virtual and erect.

Conceptual Questions

1. What is a lens? How many types of lenses are there?
A lens is a transparent optical device bounded by two curved surfaces that refracts light. There are two main types of lenses:

• Convex lens
• Concave lens

2. Describe (a) a convex lens, (b) a concave lens.
(a) Convex lens: It is thicker at the centre and thinner at the edges. It converges parallel rays of light to a point called the focus.
(b) Concave lens: It is thinner at the centre and thicker at the edges. It diverges parallel rays of light.
3. Distinguish between a convex lens and a concave lens.

Convex Lens	Concave Lens
Thicker at the centre	Thinner at the centre
Converges light rays	Diverges light rays
Can form real and virtual images	Always forms virtual image

4. Why is a convex lens called a converging lens?
A convex lens bends parallel rays of light towards a single point called the principal focus. Therefore it is called a converging lens.
5. Why is a concave lens called a diverging lens?
A concave lens spreads or diverges parallel rays of light after refraction. Hence it is called a diverging lens.
6. A thin lens has a focal length \(f=-12\) cm. Is it convex or concave?
Negative focal length indicates a concave lens.
7. If the image formed by a lens is always diminished and erect, what is the nature of the lens?
Such images are produced only by a concave lens.
8. For what position of an object is a virtual image formed by a convex lens?
When the object is placed between the optical centre and the focus (distance less than f).
9. Where should an object be placed so that a real and inverted image of the same size is obtained using a convex lens?
The object should be placed at 2f.
10. If the image formed by a convex lens is the same size as the object, where is the image formed?
The image is formed at 2f on the opposite side of the lens.
11. A 1 cm high object is placed at a distance of 2f from a convex lens. Find the height of the image.
At object distance = 2f, image is same size but inverted. \[ h_i = -1 \text{ cm} \] Negative sign shows the image is inverted.

Numerical Problems

1. Object height = 5 cm, object distance = 10 cm, focal length = 6 cm \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] \[ \frac{1}{6}=\frac{1}{v}-\frac{1}{(-10)} \] \[ \frac{1}{6}=\frac{1}{v}+\frac{1}{10} \] \[ \frac{1}{v}=\frac{1}{6}-\frac{1}{10} \] \[ v=15\,cm \] Magnification: \[ m=\frac{v}{u}=\frac{15}{-10}=-1.5 \] \[ h_i=m\times h_o=-1.5\times5=-7.5\,cm \] Answer: Image at 15 cm, real, inverted, magnified, height = 7.5 cm.
2. Focal length = 15 cm, object distance = 30 cm \[ \frac{1}{15}=\frac{1}{v}-\frac{1}{(-30)} \] \[ \frac{1}{15}=\frac{1}{v}+\frac{1}{30} \] \[ v=30\,cm \] Magnification: \[ m=\frac{30}{-30}=-1 \] Answer: Image at 30 cm, real, inverted, same size.
3. Camera lens focal length = 10 cm (a) Object at infinity \[ v=f=10\,cm \] (b) Object at 20 cm \[ \frac{1}{10}=\frac{1}{v}-\frac{1}{(-20)} \] \[ \frac{1}{10}=\frac{1}{v}+\frac{1}{20} \] \[ v=20\,cm \] Answer: (a) 10 cm (b) 20 cm
4. Object height = 4 cm, object distance = 15 cm, focal length = 5 cm \[ \frac{1}{5}=\frac{1}{v}-\frac{1}{(-15)} \] \[ \frac{1}{5}=\frac{1}{v}+\frac{1}{15} \] \[ v=7.5\,cm \] Magnification: \[ m=\frac{7.5}{-15}=-0.5 \] \[ h_i=-0.5\times4=-2\,cm \] Answer: Image at 7.5 cm, height = 2 cm, real and inverted.

Question 17

Given frequency distribution:

\(x_i\)	22	23	24	25	26	27	28
\(f_i\)	5	12	9	\(x\)	20	14	22

Mean of the data = \(25.78\) Formula: \[ \bar{x}=\frac{\sum f_i x_i}{\sum f_i} \] Sum of frequencies: \[ \sum f_i = 5+12+9+x+20+14+22 = 82+x \] Calculation of \(f_i x_i\):

\(x_i\)	\(f_i\)	\(f_i x_i\)
22	5	110
23	12	276
24	9	216
25	\(x\)	\(25x\)
26	20	520
27	14	378
28	22	616

\[ \sum f_i x_i = 2116 + 25x \] Substitute into mean formula: \[ 25.78=\frac{2116+25x}{82+x} \] \[ 25.78(82+x)=2116+25x \] \[ 2113.96+25.78x=2116+25x \] \[ 0.78x=2.04 \] \[ x\approx2.6 \] Frequency must be a whole number. \[ x=3 \] Answer: The number 25 appeared 3 times.

Question 18

Weights of 12 boys (kg): \[ 39, 51, 38, 34, 29, 31, 42, 40, 52, 44, 41, 35 \]

(a) Range

Maximum weight = 52 Minimum weight = 29 \[ \text{Range} = 52-29 = 23 \] Range = 23 kg

(b) Mean weight

\[ \text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}} \] Sum of weights: \[ 39+51+38+34+29+31+42+40+52+44+41+35 = 476 \] \[ \text{Mean}=\frac{476}{12}=39.67 \] Mean weight ≈ 39.67 kg

(c) Weight of the fattest boy

Largest value in the data: \[ 52 \text{ kg} \] Answer: 52 kg

(d) Boys having weight more than mean

Mean ≈ 39.67 Weights greater than mean: \[ 51,\ 42,\ 40,\ 52,\ 44,\ 41 \] Number of such boys: \[ 6 \] Answer: 6 boys

1. Ray travelling from rarer medium to denser medium

When a ray of light travels from a rarer medium (air) to a denser medium (glass or water), it bends towards the normal.

2. Ray striking surface at right angle

If a ray strikes the boundary perpendicular to the surface (angle of incidence = 0°), it passes straight without bending. Answer: The ray travels in the same straight line.

3. Mathematical equation of Snell’s Law

Snell’s law states: \[ n_1 \sin i = n_2 \sin r \] Where: – \( n_1 \) = refractive index of first medium – \( n_2 \) = refractive index of second medium – \( i \) = angle of incidence – \( r \) = angle of refraction

4. Speed of light in different media

Speed of light decreases as refractive index increases. Order of refractive index: Air < Water < Glass < Diamond Therefore: Highest speed: Water Lowest speed: Diamond

5. Refraction through a glass slab

In a glass slab the emergent ray becomes parallel to the incident ray but shows a lateral displacement.

6. Value of \( n_{21} \times n_{12} \)

\[ n_{21} \times n_{12} = 1 \]

7. Angle of refraction when angle of incidence is 0°

If angle of incidence = 0°, \[ r = 0^\circ \] The ray travels straight without bending.