Refraction through a Lens

Class 10 Physics • Chapter 05 (Deep Detail)

1. Types of Lenses

Convex Lens (Converging): Thicker at the middle, thinner at edges. Converges light rays to a point (Focus).

Concave Lens (Diverging): Thinner at the middle, thicker at edges. Diverges light rays.

Technical Terms:

Optical Centre (O): Point on principal axis through which ray passes undeviated.
Principal Focus (F): Point where parallel rays meet (Convex) or appear to diverge from (Concave).
Focal Length ($f$): Distance between Optical Centre and Principal Focus.

Practice Q1: Optical Centre

Q: A ray passes through the optical centre of a lens. What is its deviation?

Ans: Zero. It passes undeviated.

Practice Q2: Lens Action

CONCEPTUAL Explain the action of a convex lens on a parallel beam of light.

Ans: Converging action. It bends the parallel rays towards the principal axis to meet at the Focus.

2. Image Formation (Convex Lens)

[IMAGE PLACEHOLDER: LENS RAY DIAGRAMS]

Grid of 6 diagrams showing image formation by Convex Lens for different object positions. Use arrows for rays. Mark O, F1, 2F1, F2, 2F2.

Rules:

Ray parallel to axis $\to$ Passes through Focus ($F_2$).
Ray through Optical Centre $\to$ Goes undeviated.
Ray through Focus $\to$ Becomes parallel to axis.

Practice Q3: Ray Tracing

[IMAGE PLACEHOLDER: CONVEX LENS RAY RULES]

Convex lens. Ray 1 parallel to axis hits lens, goes through F2. Ray 2 through Optical center goes straight. Intersection is Image. (Invert in dark mode).

CONCEPTUAL Complete the path of a ray of light passing through the Focus ($F_1$) of a convex lens.

Ans: After refraction, the ray becomes parallel to the principal axis.

Cases Summary (Convex):

Object Position	Image Position	Nature
At Infinity	At Focus ($F_2$)	Real, Inverted, Point size
Beyond 2F$_1$	Between F$_2$ and 2F$_2$	Real, Inverted, Diminished
At 2F$_1$	At 2F$_2$	Real, Inverted, Same size
Between F$_1$ and 2F$_1$	Beyond 2F$_2$	Real, Inverted, Magnified
At F$_1$	At Infinity	Real, Inverted, Highly Magnified
Between F$_1$ and O	Same side (Behind Object)	Virtual, Erect, Magnified

MAGNIFYING GLASS The case where object is between $F$ and $O$ is used in a **Simple Microscope** (Magnifying Glass) and for reading small print.

Practice Q4: Ray Diagram Logic

CONCEPTUAL To get a real, inverted and magnified image using a convex lens, where should the object be placed?

Ans: Between $F_1$ and $2F_1$. (Image forms beyond $2F_2$).

Practice Q5: Magnifying Glass

THINKING To use a convex lens as a magnifying glass, where should the object be placed?

Solution: The object must be placed between the Optical Centre and the Focus ($u < f$) of the lens. This creates a virtual, erect, and magnified image.

3. Concave Lens

Always forms a Virtual, Erect, and Diminished image, located between Focus and Optical Centre on the same side as the object.

Practice Q6: Concave Image

[IMAGE PLACEHOLDER: CONCAVE LENS DIVERGING]

Concave lens. Object at 2F. Rays diverge. Trace backward to meet at virtual F. Virtual, Erect, Diminished image. (Invert in dark mode).

REASONING Can a concave lens ever form a real image? Explain.

Ans: No, for a real object, a concave lens always diverges light rays. They never actually meet, only appear to meet when produced backwards. Hence, it always forms a Virtual image.

4. Sign Convention

Light travels left to right. Optical Centre is Origin (0,0).
Direction of incident light: Positive (+ve).
Opposite to incident light: Negative (-ve).
Upwards (Height): Positive. Downwards: Negative.

Focal Length: Convex = Positive (+), Concave = Negative (-).

5. Lens Formula & Magnification

LENS FORMULA

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

Where $u$ = object distance (always -ve), $v$ = image distance, $f$ = focal length.

MAGNIFICATION (m)

$$ m = \frac{h_i}{h_o} = \frac{v}{u} $$

For Real image: $m$ is negative. For Virtual image: $m$ is positive.

Practice Q7: Projector Setup

APPLICATION A slide projector has to project a 100 times magnified image on a screen 5m away. What lens should be used?

Solution:

1. Image is Real (on screen), so $m = -100$. $v = +500 \text{ cm}$.

2. $m = v/u \implies -100 = 500/u$

$\implies u = -5 \text{ cm}$.

3. $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{500} - \frac{1}{-5} = \frac{1}{500} + \frac{100}{500} = \frac{101}{500}$.

$f \approx 4.95 \text{ cm}$ (Convex Lens).

Practice Q8: Finding Image Position

NUMERICAL An object is placed 20 cm from a convex lens of focal length 15 cm. Find image position.

Solution:

$u = -20$, $f = +15$.

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{15} - \frac{1}{20} = \frac{4-3}{60} = \frac{1}{60}$.

$v = +60 \text{ cm}$ (Real image on other side).

Practice Q9: Concave Lens

NUMERICAL An object is placed 30 cm from a concave lens of focal length 15 cm. Find image distance.

Solution:

$u = -30$, $f = -15$ (Concave is negative).

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-15} - \frac{1}{30} = \frac{-2-1}{30} = \frac{-3}{30} = - \frac{1}{10}$.

$v = -10 \text{ cm}$ (Virtual image on same side).

Practice Q10: Lens Formula

NUMERICAL An object is placed at a distance of 24 cm from a convex lens of focal length 8 cm. Find the position and nature of the image.

Solution: $u = -24, f = +8$.

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{8} - \frac{1}{24}$

$= \frac{3-1}{24} = \frac{1}{12}$.

$v = +12 \text{ cm}$. Real and Inverted.

Practice Q11: Finding Object Position (Mag -3)

NUMERICAL A convex lens produces a real image 3 times the size of the object. If focal length is 10 cm, find object position.

Solution: Real image $\implies m = -3 \implies v = -3u$.

$\frac{1}{-3u} - \frac{1}{u} = \frac{1}{10}$

$\implies \frac{-4}{3u} = \frac{1}{10}$

$\implies 3u = -40 \implies u = -13.33 \text{ cm}$.

6. Power of a Lens (P)

The ability of a lens to converge or diverge light rays.

$$ P = \frac{1}{f (\text{in metres})} $$

SI Unit: Dioptre (D).

Convex (+D), Concave (-D). (e.g., Spectacle power -2.5D means Concave lens).

Practice Q12: Lens Combination

NUMERICAL Two lenses of power +2.5 D and -1.5 D are placed in contact. Find the focal length of the combination.

Solution:

Net Power $P = P_1 + P_2 = +2.5 - 1.5 = +1.0 \text{ D}$.

Focal Length $f = 1/P = 1/1 = 1 \text{ m} = 100 \text{ cm}$.

System behaves as a Convex lens.

Practice Q13: Basic Power Calc

Q: Find the power of a convex lens of focal length 25 cm.

Ans: $f = 0.25 \text{ m}$. $P = 1/0.25 = +4 \text{ D}$.

Combination of Lenses

If two lenses of power $P_1$ and $P_2$ are placed in contact, the net power $P$ is:

$$ P = P_1 + P_2 $$

Practice Q14: Power Identification

CONCEPTUAL A lens has a power of -2.0 D. Identify the type of lens and its focal length.

Solution: Negative power $\implies$ Concave Lens.

$f = 1/P = 1/-2 = -0.5 \text{ m} = -50 \text{ cm}$.