Light – Reflection and Refraction: NCERT Solutions for Class 10 Science Chapter 9

The principal focus of a concave mirror is the point on the principal axis where rays of light that are parallel to the principal axis actually converge after reflection from the mirror. The distance between the pole of the mirror and the principal focus is called the focal length (f).

For a concave mirror, the principal focus lies in front of the mirror.

The focal length f of a spherical mirror is half its radius of curvature R, regardless of whether the mirror is concave or convex: \( f = R/2 \).

\( f = \frac{20 \text{ cm}}{2} = 10 \text{ cm} \) Answer: The focal length is 10 cm.

A concave mirror can produce an erect and enlarged image when the object is placed between its pole and principal focus (i.e., very close to the mirror). In this position, the image is virtual, erect and larger than the object. This is why concave mirrors are used as shaving mirrors or by dentists to see enlarged images of teeth.

Convex mirrors are preferred as rear-view mirrors for two reasons:

• They always form an erect image (virtual), which is essential for the driver to see traffic in the correct orientation.
• They have a wider field of view because the curved-outward surface enables the mirror to capture light from a larger area than a plane or concave mirror of the same size would.

Even though the image is smaller than the object, the advantage of seeing a broader area behind the vehicle makes convex mirrors the standard choice for safe driving.

The radius of curvature is given as 32 cm. For any spherical mirror, the focal length is half the radius of curvature. For a convex mirror, the focal length is positive (by sign convention).

\( f = \frac{R}{2} = \frac{32 \text{ cm}}{2} = +16 \text{ cm} \) Answer: The focal length is +16 cm.

The object is placed in front of the mirror, so \( u = -10 \text{ cm} \). The image is real and three times magnified. For a real image, the magnification is negative (inverted). Therefore \( m = -3 \).

Using the magnification formula for mirrors: \( m = -\frac{v}{u} \).

\( -3 = -\frac{v}{(-10)} \) \( -3 = -\frac{v}{-10} = -\left( -\frac{v}{10} \right) = \frac{v}{10} \) \( v = -30 \text{ cm} \) Answer: The image is located 30 cm in front of the mirror (real image).

When light travels from a rarer medium (air) to a denser medium (water), it slows down. The decrease in speed causes the light ray to bend towards the normal. This is because the decrease in speed is more pronounced in the denser medium, and the ray must change direction to satisfy Snell’s law (n = sin i / sin r).

The angle of refraction is smaller than the angle of incidence, which means the ray bends towards the normal.

The refractive index n of a medium is defined as the ratio of the speed of light in vacuum c to the speed of light in that medium v: \( n = \frac{c}{v} \).

\( v = \frac{c}{n} = \frac{3 \times 10^8 \text{ m/s}}{1.50} = 2 \times 10^8 \text{ m/s} \) Answer: The speed of light in glass is \( 2 \times 10^8 \text{ m/s} \).

Table 9.3 lists the absolute refractive indices of various materials. The medium with the highest refractive index has the highest optical density, and the one with the lowest refractive index has the lowest optical density.

From the table, the medium with the highest optical density is diamond (refractive index 2.42). The medium with the lowest optical density is air (refractive index 1.0003).

From Table 9.3, the refractive indices are:

• Kerosene: 1.44
• Turpentine oil: 1.47
• Water: 1.33

Light travels fastest in the medium with the lowest refractive index because speed v = c/n. Among the three, water has the lowest refractive index (1.33), so light travels fastest in water.

The refractive index of diamond is 2.42. This means that the speed of light in diamond is \( 1/2.42 \) times the speed of light in vacuum (or air). In other words, light travels 2.42 times slower in diamond than in vacuum. Mathematically, \( n_{\text{diamond}} = \frac{c}{v_{\text{diamond}}} = 2.42 \), so \( v_{\text{diamond}} = \frac{c}{2.42} \).

One dioptre (1 D) is the power of a lens whose focal length is 1 metre. The power of a lens is the reciprocal of its focal length in metres: \( P = 1/f \). So if \( f = 1 \text{ m} \), then \( P = 1 \text{ D} \). The unit dioptre is equivalent to \( \text{m}^{-1} \).

When a convex lens produces a real image of the same size as the object, the object must be placed at twice the focal length (2F) from the lens. At this position, the image is also formed at 2F on the other side, and the image distance equals the object distance.

Given: image distance \( v = +50 \text{ cm} \) (real image is on the opposite side, so positive). At 2F, \( v = 2f \), so \( 2f = 50 \text{ cm} \) → \( f = 25 \text{ cm} = 0.25 \text{ m} \).

The object distance is also \( u = -50 \text{ cm} \) (object is left of the lens).

Power of the lens: \( P = \frac{1}{f} = \frac{1}{0.25 \text{ m}} = +4 \text{ D} \).

Answer: The needle is placed 50 cm in front of the lens. The power of the lens is +4 D.

For a concave lens, the focal length is negative by sign convention. Given focal length \( f = -2 \text{ m} \).

\( P = \frac{1}{f} = \frac{1}{-2 \text{ m}} = -0.5 \text{ D} \) Answer: The power of the concave lens is -0.5 D.

A lens must be made of a transparent material so that light can pass through it and undergo refraction. Among the options, clay is opaque – it does not allow light to pass through. Therefore, clay cannot be used to make a lens. Water, glass and plastic are all transparent and can be shaped into lenses.

Correct answer: (d) Clay

According to Table 9.1 (Image formation by a concave mirror), a virtual, erect and enlarged image is formed when the object is placed between the pole (P) and the principal focus (F) of the concave mirror. In this position, the image is behind the mirror, virtual, erect and larger than the object.

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

For a convex lens (converging lens), a real image of the same size as the object is formed when the object is placed at twice the focal length (2F). At this position, the image is also formed at 2F on the other side, and the image size equals the object size.

If the object is at the principal focus, the image is at infinity. If the object is at infinity, the image is at the focus. If the object is between the optical centre and the focus, the image is virtual and enlarged.

Correct answer: (b) At twice the focal length

According to the New Cartesian Sign Convention:

• A concave mirror has a negative focal length.
• A convex mirror has a positive focal length.
• A convex lens has a positive focal length.
• A concave lens has a negative focal length.

Both have focal length -15 cm, meaning the mirror is concave and the lens is concave.

Correct answer: (a) both concave.

A mirror that always forms an erect image regardless of the object distance can be either a plane mirror or a convex mirror.

• A plane mirror always forms an erect, virtual image of the same size.
• A convex mirror always forms an erect, virtual, diminished image.
• A concave mirror can form an erect image only when the object is placed between the pole and focus; otherwise it forms an inverted image.

Therefore, the mirror must be either plane or convex.

Correct answer: (d) either plane or convex.

To read small letters, you need a magnifying glass, which is a convex lens. The magnification produced by a convex lens when the object is within the focal length is given by \( m = 1 + D/f \), where D is the near point distance (approximately 25 cm). A shorter focal length gives higher magnification.

Therefore, a convex lens of focal length 5 cm will magnify the letters more than one of focal length 50 cm. A concave lens always produces a diminished image, so it is not suitable.

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

For a concave mirror, an erect image is obtained only when the object is placed between the pole (P) and the principal focus (F). Since the focal length is 15 cm, the object must be placed at a distance less than 15 cm from the mirror (i.e., between 0 and 15 cm from the pole).

• Range of object distance: 0 < u < 15 cm (i.e., between the pole and the focus).
• Nature of the image: Virtual and erect.
• Size of the image: Larger than the object (enlarged).

Ray diagram description (refer to Fig. 9.7(f) in the textbook):

Using the two-ray construction:

1. A ray parallel to the principal axis reflects and passes through the focus F.
2. A ray striking the pole reflects symmetrically, obeying the law of reflection.
3. The reflected rays diverge; extend them backwards behind the mirror – they meet at a point between P and F behind the mirror. This is the virtual image.

Part (a): Headlights of a car Mirror used: Concave mirror.

Reason: When a light source is placed at the principal focus of a concave mirror, the reflected rays become a parallel beam of light. This allows the car headlights to throw a strong, parallel beam far ahead on the road.

Part (b): Side/rear-view mirror of a vehicle Mirror used: Convex mirror.

Reason: Convex mirrors always produce an erect image (virtual) and have a wider field of view. This enables the driver to see more of the traffic behind than a plane mirror of the same size would show. The diminished image is acceptable because the key requirement is the larger area visible.

Part (c): Solar furnace Mirror used: Large concave mirror.

Reason: A concave mirror converges sunlight to its focus. By placing the furnace at the focus, a large amount of solar energy is concentrated at a small area, producing intense heat. This is the principle behind solar furnaces.

Answer: Yes, the lens will still produce a complete image of the object, but the image will be less bright (dimmer) than when the full lens is used.

Reason: Every point of an object sends rays of light in all directions, and these rays pass through every part of the lens. Even if half the lens is covered, the uncovered half still receives rays from all points of the object. The lens still refracts these rays to form a complete image at the same position.

However, because only half the area is available to collect light, the amount of light reaching the image is reduced, making the image dimmer.

Experimental verification:

1. Take a convex lens and form a sharp image of a candle flame on a screen.
2. Cover the upper half of the lens with a black paper (or any opaque material).
3. Observe the image on the screen. You will still see a complete image of the candle flame, but it will be fainter.
4. If you cover the entire lens, no image is formed.

Conclusion: The covered half does not block the image of any part of the object; it only reduces the intensity of the image.

Given: object height \( h = +5 \text{ cm} \), object distance \( u = -25 \text{ cm} \) (convention: left of lens), focal length of converging (convex) lens \( f = +10 \text{ cm} \).

Step 1: Use the lens formula.

\( \frac{1}{v} – \frac{1}{u} = \frac{1}{f} \) \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{(-25)} = \frac{1}{10} – \frac{1}{25} \) \( \frac{1}{v} = \frac{5 – 2}{50} = \frac{3}{50} \) \( v = \frac{50}{3} \approx +16.67 \text{ cm} \) The positive sign of v indicates that the image is formed on the opposite side of the lens (real image).

Step 2: Magnification.

\( m = \frac{v}{u} = \frac{+16.67}{-25} = -\frac{2}{3} \approx -0.67 \) Step 3: Image height.

\( h’ = m \times h = -\frac{2}{3} \times 5 = -\frac{10}{3} \approx -3.33 \text{ cm} \) Final answer:

• Position: 16.7 cm on the other side of the lens (real image).
• Nature: Real and inverted.
• Size: 3.33 cm (diminished, about 2/3 of the object size).

Given: concave lens, so \( f = -15 \text{ cm} \). The image is virtual and formed on the same side as the object, so \( v = -10 \text{ cm} \).

Step 1: Lens formula.

\( \frac{1}{v} – \frac{1}{u} = \frac{1}{f} \) \( \frac{1}{u} = \frac{1}{v} – \frac{1}{f} = \frac{1}{-10} – \frac{1}{(-15)} = -\frac{1}{10} + \frac{1}{15} \) \( \frac{1}{u} = \frac{-3 + 2}{30} = -\frac{1}{30} \) \( u = -30 \text{ cm} \) Step 2: Magnification.

\( m = \frac{v}{u} = \frac{-10}{-30} = +\frac{1}{3} \approx 0.33 \) The positive sign indicates the image is erect and virtual. Size is one-third of the object.

Final answer: The object is placed 30 cm from the lens (on the same side). The image is virtual, erect and diminished.

Given: convex mirror, \( f = +15 \text{ cm} \); object distance \( u = -10 \text{ cm} \).

Step 1: Use the mirror formula (since it’s a mirror).

\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \) \( \frac{1}{v} = \frac{1}{f} – \frac{1}{u} = \frac{1}{15} – \frac{1}{(-10)} = \frac{1}{15} + \frac{1}{10} \) \( \frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6} \) \( v = +6 \text{ cm} \) Step 2: Magnification.

\( m = -\frac{v}{u} = -\frac{+6}{-10} = +0.6 \) The positive sign indicates the image is virtual and erect. Magnification less than 1 means it is diminished.

Final answer: The image is formed 6 cm behind the mirror (virtual, erect, diminished).

Magnification \( m = +1 \) means:

• The image is the same size as the object (magnitude 1).
• The image is erect (positive sign).
• The image is virtual (since the magnification is positive, the image is virtual).

This matches the characteristics of a plane mirror: it always produces a virtual, erect image of the same size as the object.

Given: convex mirror, radius of curvature \( R = +30 \text{ cm} \) (positive for convex). Focal length \( f = R/2 = +15 \text{ cm} \). Object distance \( u = -20 \text{ cm} \). Object height \( h = +5.0 \text{ cm} \).

Step 1: Mirror formula.

\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \) \( \frac{1}{v} = \frac{1}{f} – \frac{1}{u} = \frac{1}{15} – \frac{1}{(-20)} = \frac{1}{15} + \frac{1}{20} \) \( \frac{1}{v} = \frac{4 + 3}{60} = \frac{7}{60} \) \( v = \frac{60}{7} \approx +8.57 \text{ cm} \) Step 2: Magnification.

\( m = -\frac{v}{u} = -\frac{+8.57}{-20} = +0.4285 \approx +0.43 \) Step 3: Image height.

\( h’ = m \times h = 0.4285 \times 5.0 = +2.14 \text{ cm} \) Final answer: The image is formed 8.6 cm behind the mirror.

It is virtual, erect and diminished (size ≈ 2.1 cm).

Given: concave mirror, \( f = -18 \text{ cm} \), object distance \( u = -27 \text{ cm} \), object height \( h = +7.0 \text{ cm} \).

Step 1: Mirror formula.

\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \) \( \frac{1}{v} = \frac{1}{f} – \frac{1}{u} = \frac{1}{-18} – \frac{1}{(-27)} = -\frac{1}{18} + \frac{1}{27} \) \( \frac{1}{v} = \frac{-3 + 2}{54} = -\frac{1}{54} \) \( v = -54 \text{ cm} \) The negative sign indicates that the image is real and formed on the same side as the object. Thus, a screen should be placed 54 cm in front of the mirror.

Step 2: Magnification.

\( m = -\frac{v}{u} = -\frac{-54}{-27} = -\frac{54}{27} = -2 \) Step 3: Image height.

\( h’ = m \times h = -2 \times 7.0 = -14 \text{ cm} \) Final answer: The screen should be placed 54 cm from the mirror.

The image is real, inverted, and enlarged (14 cm tall).

The power P of a lens is related to its focal length f (in metres) by \( P = 1/f \).

\( f = \frac{1}{P} = \frac{1}{-2.0 \text{ D}} = -0.5 \text{ m} \) The negative focal length indicates that the lens is a concave lens (diverging lens).

Answer: Focal length = -0.5 m, concave lens.

Given power \( P = +1.5 \text{ D} \).

\( f = \frac{1}{P} = \frac{1}{+1.5} = +0.6667 \text{ m} \approx +0.67 \text{ m} \) The positive focal length indicates that the lens is a convex lens (converging lens).

Answer: Focal length = +0.67 m, converging lens (convex).