Light - Reflection and Refraction

Definition: 1 dioptre is the power of a lens whose focal length is 1 metre.

Explanation:
The power of a lens is defined as the reciprocal of its focal length in metres:
$$P = \frac{1}{f(\text{in metres})}$$

When $f = 1\,\text{m}$,
$$P = \frac{1}{1\,\text{m}} = 1\,\text{D}$$

Thus, 1 dioptre (1 D) is the power of a lens of focal length 1 metre. Its SI unit is $\text{m}^{-1}$.

Given:
- Image distance, $v = +50\,\text{cm}$ (real and inverted, so positive by sign convention)
- Image size = Object size, so magnification $m = -1$ (real and inverted image of same size)

Step 1: Find object distance using magnification.
$$m = \frac{v}{u}$$
$$-1 = \frac{+50}{u}$$
$$u = -50\,\text{cm}$$

So the needle is placed 50 cm in front of the lens (i.e., at the centre of curvature / at $2f$).

Step 2: Find focal length using lens formula.
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{+50} - \frac{1}{-50} = \frac{1}{f}$$
$$\frac{1}{50} + \frac{1}{50} = \frac{1}{f}$$
$$\frac{2}{50} = \frac{1}{f}$$
$$f = 25\,\text{cm} = 0.25\,\text{m}$$

Step 3: Find power of the lens.
$$P = \frac{1}{f} = \frac{1}{0.25\,\text{m}} = +4\,\text{D}$$

Result: The needle is placed 50 cm in front of the convex lens (at twice the focal length), and the power of the lens is $\mathbf{+4\,D}$.

Given:
- Focal length of concave lens, $f = -2\,\text{m}$ (negative for concave lens)

Formula:
$$P = \frac{1}{f}$$

Calculation:
$$P = \frac{1}{-2\,\text{m}} = -0.5\,\text{D}$$

Result: The power of the concave lens is $\mathbf{-0.5\,D}$.

Correct Answer: (d) Clay

Justification: A lens works on the principle of refraction of light. For refraction to occur, the material must be transparent (light must pass through it). Clay is an opaque material — light cannot pass through it — so it cannot be used to make a lens. Water, glass, and plastic are all transparent and can be used to make lenses.

Correct Answer: (d) Between the pole of the mirror and its principal focus.

Justification: A concave mirror produces a virtual, erect, and magnified image only when the object is placed between the pole (P) and the principal focus (F) of the mirror. In all other positions, a concave mirror forms a real and inverted image.

Correct Answer: (b) At twice the focal length.

Justification: When an object is placed at $2f$ (twice the focal length) in front of a convex lens, the image is formed at $2f$ on the other side. The image is real, inverted, and of the same size as the object (magnification $= -1$). This is the only position where image size equals object size for a convex lens.

Correct Answer: (a) both concave.

Justification: By the New Cartesian Sign Convention:
- A concave mirror has a negative focal length (focus is in front of the mirror, on the same side as the object).
- A concave lens (diverging lens) also has a negative focal length.

Since both have $f = -15\,\text{cm}$ (negative), both the mirror and the lens are concave.

Correct Answer: (d) either plane or convex.

Justification:
- A plane mirror always forms a virtual, erect image regardless of the distance of the object.
- A convex mirror always forms a virtual, erect (and diminished) image for any position of the object.
- A concave mirror forms an erect image only when the object is between the pole and focus; beyond that it forms inverted images.

Therefore, the mirror is either plane or convex.

Correct Answer: (c) A convex lens of focal length 5 cm.

Justification: To read small letters, we need a magnifying glass, which is a convex (converging) lens. A convex lens produces a magnified, virtual, and erect image when the object is within the focal length. A shorter focal length gives greater magnifying power ($P = 1/f$, so smaller $f$ means larger $P$). Hence, a convex lens of focal length 5 cm provides the greatest magnification among the given options.

Given: Focal length of concave mirror, $f = 15\,\text{cm}$

Condition for erect image with a concave mirror:
A concave mirror produces a virtual, erect, and magnified image only when the object is placed between the pole (P) and the principal focus (F).

Range of object distance:
0 < u < 15\,\text{cm}
(i.e., the object must be placed between the pole and the focus, less than 15 cm from the mirror)

Nature of the image:
- Virtual
- Erect
- Magnified (larger than the object)

Ray Diagram Description:
Place the object between P and F. Draw two rays from the top of the object:
1. A ray parallel to the principal axis — after reflection, it passes through (or appears to come from) the focus F.
2. A ray directed towards the centre of curvature C — it reflects back along the same path.

The reflected rays diverge and when extended behind the mirror, they meet to form a virtual, erect, and enlarged image behind the mirror.

Conclusion: The object should be placed at a distance less than 15 cm from the concave mirror. The image formed is virtual, erect, and larger than the object.

Answers:

(a) Headlights of a car — Concave Mirror

Reason: A concave mirror converges light rays. When a bulb (light source) is placed at the focus of a concave mirror, the reflected rays emerge as a powerful parallel beam of light. This provides a strong, focused beam of light for illuminating the road ahead.

(b) Side/rear-view mirror of a vehicle — Convex Mirror

Reason: A convex mirror always forms a virtual, erect, and diminished image. It has a wider field of view compared to a plane or concave mirror of the same size. This allows the driver to see a larger area of the traffic behind the vehicle, making it safer for driving.

(c) Solar furnace — Concave Mirror

Reason: A large concave mirror (parabolic reflector) converges all the incident parallel rays of sunlight at its focus. This concentrates a large amount of solar energy at one point (the focus), producing very high temperatures, which can be used to heat substances in a solar furnace.

Answer: Yes, the lens will still produce a complete image of the object.

Explanation:
When half of a convex lens is covered with black paper, the uncovered half still refracts light rays coming from the object. Every part of a lens forms a complete image of the object. The covered half simply does not allow light to pass through it, but the remaining half still converges the light rays to form a complete image.

Observation:
- The image formed is complete (not half).
- However, the image is less bright (dimmer) compared to the image formed by the full lens, because only half the amount of light passes through the lens.

Experimental Verification:
Place a lighted candle in front of a convex lens and obtain its sharp image on a screen. Now cover the lower half of the lens with black paper. Observe that a complete image of the candle is still formed on the screen, but it is less bright than before.

Conclusion: Covering half the lens reduces the intensity (brightness) of the image but does not make the image incomplete. A complete image is still formed.

Given:
- Object height, $h_o = +5\,\text{cm}$
- Object distance, $u = -25\,\text{cm}$ (object is on the left side)
- Focal length, $f = +10\,\text{cm}$ (converging/convex lens)

Step 1: Apply the lens formula.
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} - \frac{1}{-25} = \frac{1}{10}$$
$$\frac{1}{v} + \frac{1}{25} = \frac{1}{10}$$
$$\frac{1}{v} = \frac{1}{10} - \frac{1}{25}$$
$$\frac{1}{v} = \frac{5 - 2}{50} = \frac{3}{50}$$
$$v = +\frac{50}{3} \approx +16.67\,\text{cm}$$

Step 2: Find magnification.
$$m = \frac{v}{u} = \frac{+50/3}{-25} = \frac{50}{3 \times (-25)} = -\frac{2}{3}$$

Step 3: Find image height.
$$h_i = m \times h_o = -\frac{2}{3} \times 5 = -\frac{10}{3} \approx -3.33\,\text{cm}$$

Results:
- Position of image: $v \approx +16.67\,\text{cm}$ on the other side of the lens.
- Size of image: $|h_i| \approx 3.33\,\text{cm}$ (smaller than the object)
- Nature of image: Real, inverted, and diminished.

Ray Diagram Description: The object is beyond $f$ but within $2f$... actually $u = 25\,\text{cm}$ and $2f = 20\,\text{cm}$, so the object is beyond $2f$. The image forms between $f$ and $2f$ on the other side, real, inverted, and diminished.

Given:
- Focal length of concave lens, $f = -15\,\text{cm}$ (negative for concave lens)
- Image distance, $v = -10\,\text{cm}$ (image formed on the same side as the object for a concave lens — virtual image)

Step 1: Apply the lens formula.
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{-10} - \frac{1}{u} = \frac{1}{-15}$$
$$-\frac{1}{u} = \frac{1}{-15} - \frac{1}{-10}$$
$$-\frac{1}{u} = -\frac{1}{15} + \frac{1}{10}$$
$$-\frac{1}{u} = \frac{-2 + 3}{30} = \frac{1}{30}$$
$$\frac{1}{u} = -\frac{1}{30}$$
$$u = -30\,\text{cm}$$

Result: The object is placed 30 cm in front of the concave lens (on the same side as the incoming light).

Ray Diagram Description: The object is placed 30 cm to the left of the concave lens. Two rays from the top of the object:
1. A ray parallel to the principal axis — after refraction, it appears to diverge from the focus $F_1$ on the same side as the object.
2. A ray directed towards the optical centre — passes straight through without bending.

The two refracted rays diverge; when extended backwards, they meet at $v = -10\,\text{cm}$ on the same side as the object, forming a virtual, erect, and diminished image.

Given:
- Object distance, $u = -10\,\text{cm}$ (object in front of mirror)
- Focal length of convex mirror, $f = +15\,\text{cm}$ (positive for convex mirror)

Step 1: Apply the mirror formula.
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} + \frac{1}{-10} = \frac{1}{+15}$$
$$\frac{1}{v} = \frac{1}{15} + \frac{1}{10}$$
$$\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$$
$$v = +6\,\text{cm}$$

Step 2: Find magnification.
$$m = -\frac{v}{u} = -\frac{+6}{-10} = +0.6$$

Results:
- Position: The image is formed 6 cm behind the convex mirror (behind the mirror, on the other side).
- Nature of image: Virtual, erect, and diminished (since $m = +0.6$, which is positive and less than 1).

Meaning of magnification $m = +1$ for a plane mirror:

The formula for magnification is:
$$m = \frac{\text{Height of image}}{\text{Height of object}} = \frac{h_i}{h_o}$$

$m = +1$ means:
1. Positive sign (+): The image is virtual and erect (same orientation as the object — not inverted).
2. Magnitude = 1: The size (height) of the image is equal to the size of the object — neither magnified nor diminished.

Conclusion: A plane mirror forms an image that is virtual, erect, and of the same size as the object. The image is formed as far behind the mirror as the object is in front of it.

Given:
- Object height, $h_o = +5.0\,\text{cm}$
- Object distance, $u = -20\,\text{cm}$
- Radius of curvature, $R = +30\,\text{cm}$ (convex mirror, so positive)
- Focal length, $f = \dfrac{R}{2} = \dfrac{+30}{2} = +15\,\text{cm}$

Step 1: Apply the mirror formula.
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} + \frac{1}{-20} = \frac{1}{+15}$$
$$\frac{1}{v} = \frac{1}{15} + \frac{1}{20}$$
$$\frac{1}{v} = \frac{4 + 3}{60} = \frac{7}{60}$$
$$v = +\frac{60}{7} \approx +8.57\,\text{cm}$$

Step 2: Find magnification.
$$m = -\frac{v}{u} = -\frac{+60/7}{-20} = \frac{60}{7 \times 20} = \frac{60}{140} = +\frac{3}{7} \approx +0.43$$

Step 3: Find image height.
$$h_i = m \times h_o = +\frac{3}{7} \times 5.0 = +\frac{15}{7} \approx +2.14\,\text{cm}$$

Results:
- Position: Image is formed $\approx 8.57\,\text{cm}$ behind the convex mirror.
- Nature: Virtual, erect, and diminished.
- Size: $\approx 2.14\,\text{cm}$ (smaller than the object).

Given:
- Object height, $h_o = +7.0\,\text{cm}$
- Object distance, $u = -27\,\text{cm}$
- Focal length of concave mirror, $f = -18\,\text{cm}$

Step 1: Apply the mirror formula.
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} + \frac{1}{-27} = \frac{1}{-18}$$
$$\frac{1}{v} = \frac{1}{-18} + \frac{1}{27}$$
$$\frac{1}{v} = \frac{-3 + 2}{54} = \frac{-1}{54}$$
$$v = -54\,\text{cm}$$

Step 2: Find magnification.
$$m = -\frac{v}{u} = -\frac{-54}{-27} = -\frac{54}{27} = -2$$

Step 3: Find image height.
$$h_i = m \times h_o = -2 \times 7.0 = -14\,\text{cm}$$

Results:
- Screen position: The screen should be placed 54 cm in front of the concave mirror (on the same side as the object).
- Size of image: $|h_i| = 14\,\text{cm}$ (twice the size of the object)
- Nature of image: Real, inverted, and magnified (enlarged).

Given:
- Power of lens, $P = -2.0\,\text{D}$

Formula:
$$P = \frac{1}{f} \implies f = \frac{1}{P}$$

Calculation:
$$f = \frac{1}{-2.0\,\text{D}} = -0.5\,\text{m}$$

Result:
- Focal length: $f = -0.5\,\text{m}$ (or $-50\,\text{cm}$)
- Type of lens: Since the focal length is negative, this is a concave (diverging) lens.

Given:
- Power of lens, $P = +1.5\,\text{D}$

Formula:
$$f = \frac{1}{P}$$

Calculation:
$$f = \frac{1}{+1.5} = \frac{1}{\frac{3}{2}} = \frac{2}{3}\,\text{m} \approx +0.67\,\text{m}$$

Result:
- Focal length: $f \approx +0.67\,\text{m}$ (or $\approx 66.7\,\text{cm}$)
- Type of lens: Since the power is positive and the focal length is positive, the prescribed lens is a converging (convex) lens.