NCERT Solutions Class 10 Science Chapter 9 Light Reflection and Refraction

A concave mirror is a converging mirror: it bends parallel light inwards. The principal axis is the straight line through the pole and the centre of curvature. When a beam of light parallel to the principal axis falls on the mirror, every reflected ray obeys the laws of reflection and crosses the axis at one common point.

• Send several rays parallel to the principal axis onto the concave mirror.
• Each ray reflects so that the angle of incidence equals the angle of reflection.
• All the reflected rays actually meet at a single point on the principal axis, called the principal focus F.
• Because the rays really pass through this point, the concave mirror's focus is a real focus, lying in front of the mirror.

Answer: The principal focus (F) of a concave mirror is the point on its principal axis where rays of light coming parallel to the principal axis actually meet after reflection from the mirror.

For a spherical mirror, the focal length is exactly half of the radius of curvature, because the principal focus lies midway between the pole and the centre of curvature: f = R/2.

• Write the relation: f = R/2.
• Substitute the given radius of curvature, R = 20 cm: f = 20/2.
• Do the arithmetic: f = 10 cm.

Answer: The focal length of the mirror is 10 cm.

The nature of the image made by a concave mirror depends on where the object is kept. Only a converging mirror can magnify, and it gives an erect image in one special position: when the object is placed between the pole and the principal focus.

• A convex mirror always gives a virtual, erect but diminished image, so it cannot enlarge.
• A plane mirror always gives a same-size image, so it cannot enlarge either.
• A concave mirror enlarges the image when the object is between the pole P and the focus F. The image is then virtual, erect and larger (the shaving-mirror position).

Answer: A concave mirror gives an erect and enlarged image when the object is placed between its pole and its principal focus.

A convex mirror is a diverging mirror. It always forms a virtual, erect and diminished image whatever the object position. Because the image is small, a single mirror can show a much larger region behind the vehicle, giving a wide field of view.

• A convex mirror always gives an erect (upright) image, so the driver sees vehicles the right way up.
• The image is always diminished, so a lot of traffic is shrunk to fit into a small mirror, widening the field of view.
• The driver can therefore see a larger area of the road behind and judge the situation more safely.

Answer: A convex mirror gives an erect, diminished image with a wide field of view, so the driver sees a large area of traffic behind in an upright image, making it the safe choice for a rear-view mirror.

For any spherical mirror the focal length is half the radius of curvature, f = R/2. For a convex mirror the principal focus lies behind the mirror, so by the New Cartesian Sign Convention both R and f are positive.

• Write the relation: f = R/2.
• Substitute R = 32 cm (positive for a convex mirror): f = +32/2.
• Do the arithmetic: f = +16 cm.

Answer: The focal length of the convex mirror is 16 cm (taken as +16 cm because the focus is behind a convex mirror).

Magnification links image distance v and object distance u: m = -v/u. A real image made by a concave mirror is inverted, so its magnification is negative. "Three times enlarged real" means m = -3. By the sign convention the object is in front, so u = -10 cm.

• Data with signs: u = -10 cm, m = -3.
• Rearrange the magnification formula: m = -v/u ⇒ v = -m×u.
• Substitute: v = -(-3) × (-10).
• Do the arithmetic: v = -30 cm.
• The negative sign means the image is in front of the mirror, on the same side as the object, which is where a real image forms.

Answer: The image is located 30 cm in front of the mirror (v = -30 cm), real and inverted.

Refraction is the bending of light when it passes from one transparent medium to another because its speed changes. The rule: going from an optically rarer medium to an optically denser medium, light slows down and bends towards the normal; denser to rarer, it speeds up and bends away.

• Air is optically rarer (light is faster) and water is optically denser (light is slower); water's refractive index (1.33) is greater than air's (1.00).
• Light therefore slows down as it crosses from air into water.
• Slowing down (rarer to denser) makes the ray bend towards the normal, so the angle of refraction is smaller than the angle of incidence.

Answer: The light ray bends towards the normal, because it passes from optically rarer air into optically denser water, where it slows down.

The refractive index of a medium tells us how much slower light travels in it compared with vacuum: n = c/v, where c is the speed in vacuum and v the speed in the medium. Rearranging gives v = c/n.

• Make v the subject: n = c/v ⇒ v = c/n.
• Substitute c = 3 × 10⁸ m s^-1 and n = 1.50: v = (3 × 10⁸) / 1.50.
• Do the arithmetic: v = 2 × 10⁸ m s^-1.

Answer: The speed of light in the glass is 2 × 10⁸ m s^-1.

Optical density is measured by the refractive index: the higher the refractive index, the higher the optical density (and the slower light travels). So we read the table and pick the largest and smallest refractive index. Table 9.3 lists, among others, diamond at n = 2.42 and air at n = 1.0003.

• The quantity that measures optical density is the refractive index n.
• The largest n is diamond, n = 2.42, so it is optically densest.
• The smallest n is air, n = 1.0003, so it is optically rarest.

Answer: Highest optical density: diamond (n = 2.42). Lowest optical density: air (n = 1.0003).

Light travels fastest in the medium with the lowest refractive index, because v = c/n: a smaller n gives a larger speed v. From Table 9.3 the refractive indices are kerosene 1.44, turpentine 1.47 and water 1.33.

• Recall: v = c/n ⇒ smaller n means larger v.
• Compare: kerosene 1.44, turpentine 1.47, water 1.33.
• The smallest refractive index is water (1.33), so light has its highest speed in water.

Answer: Light travels fastest in water, because it has the lowest refractive index (1.33) of the three, and a lower refractive index means a higher speed.

The refractive index of a medium with respect to vacuum is the ratio of the speed of light in vacuum to the speed of light in that medium: n = c/v. A value of 2.42 is this ratio for diamond.

• From the definition: n_diamond = c / v_diamond = 2.42.
• In words: the speed of light in vacuum is 2.42 times the speed of light in diamond.
• Equivalently, light travels 2.42 times slower in diamond, so v_diamond = c/2.42 ≈ 1.24 × 10⁸ m s^-1.

Answer: It means the speed of light in vacuum is 2.42 times the speed of light in diamond, i.e. light slows down by a factor of 2.42 on entering diamond (v ≈ 1.24 × 10⁸ m s^-1).

The power of a lens measures how strongly it converges or diverges light, and is the reciprocal of its focal length in metres: P = 1/f (f in metres). Its SI unit is the dioptre (D).

• Write the definition: P = 1/f, with f in metres.
• Set P = 1 D: 1 = 1/f.
• Solve: f = 1 m. So 1 dioptre is the power of a lens of focal length 1 metre.

Answer: 1 dioptre (1 D) is the power of a lens whose focal length is 1 metre.

A convex lens forms a real, inverted image of the same size as the object only when the object is at twice the focal length (2F); the image also forms at 2F on the other side. For a lens, magnification m = v/u; an equal-size, real, inverted image has m = -1.

• The image is real and on the far side, so v = +50 cm. Equal-size inverted means m = v/u = -1, so u = -v = -50 cm. The needle is 50 cm in front of the lens.
• Apply the lens formula: 1/v - 1/u = 1/f.
• Substitute: 1/50 - 1/(-50) = 1/f.
• Simplify: 1/50 + 1/50 = 2/50 = 1/25, so f = +25 cm = +0.25 m.
• Power: P = 1/f = 1/0.25 = +4 D.

Answer: The needle is placed 50 cm in front of the lens (at 2F); the focal length is 25 cm and the power is P = +4 D.

The power of a lens is the reciprocal of its focal length in metres, P = 1/f. A concave (diverging) lens has a negative focal length, so its power is negative.

• Apply the sign convention: a concave lens is diverging, so f = -2 m.
• Write the formula: P = 1/f.
• Substitute: P = 1/(-2).
• Do the arithmetic: P = -0.5 D.

Answer: The power of the concave lens is -0.5 D (the negative sign shows it is a diverging lens).

A lens works by refraction: light must pass through it and bend. So a lens can be made only from a transparent material. Any opaque material cannot form a lens.

• Water is transparent and can act as a lens (a water drop magnifies).
• Glass is transparent and is the usual lens material; plastic is also transparent and is used for cheap lenses and spectacles.
• Clay is opaque: light cannot pass through it, so it cannot refract light and cannot make a lens.

Answer: (d) Clay, because it is opaque and light cannot pass through it, so it cannot refract light to form a lens.

A concave mirror gives a virtual, erect and magnified image in only one situation: when the object lies between the pole P and the principal focus F. In every other position the image is real and inverted.

• Object beyond C, at C, or between C and F all give real, inverted images, so (a), (b) and (c) are ruled out.
• Only when the object is between P and F does the mirror form a virtual, erect, enlarged image (the shaving-mirror position).
• This matches option (d).

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

A convex lens forms a real, inverted image of the same size as the object only when the object is placed at twice the focal length (2F). The image then forms at 2F on the other side.

• Object at infinity gives a point image at F (c wrong); object at F gives an image at infinity (a wrong); object between optical centre and F gives a virtual, erect, enlarged image (d wrong).
• Object at 2F gives a real, inverted image of the same size at 2F on the other side.
• So the correct position is at twice the focal length, option (b).

Answer: (b) At twice the focal length (2F).

The sign of the focal length tells us the type of mirror or lens. A concave mirror has a negative focal length, and a concave lens also has a negative focal length. So f = -15 cm means concave for both.

• For mirrors: a concave mirror's focus is in front, so f is negative; a convex mirror's f is positive. Here f = -15 cm, so the mirror is concave.
• For lenses: a concave (diverging) lens has a negative focal length; a convex (converging) lens has a positive one. Here f = -15 cm, so the lens is concave.
• Both are concave, which is option (a).

Answer: (a) Both concave (a negative focal length means a concave mirror and a concave lens).

We need the mirror that gives an erect image for every object distance. A plane mirror always gives an erect, virtual, same-size image; a convex mirror always gives an erect, virtual, diminished image. A concave mirror gives an erect image only when the object is very close (inside the focus).

• Concave mirror: for objects beyond the focus it forms a real, inverted image, so it fails the "always erect" test. Rule out (b).
• Plane mirror: erect image at all distances. It works.
• Convex mirror: erect image at all distances. It works too.
• Since both plane and convex mirrors always give an erect image, the answer is "either plane or convex," option (d).

Answer: (d) Either plane or convex (both give an erect image for every object distance).

To read small letters we need a magnifying glass, which is a convex lens used with the object inside its focus. A concave lens always makes things smaller, so it is useless for reading. Among convex lenses, a shorter focal length gives a higher magnification.

• Reject the concave lenses (b) and (d): a concave lens always diminishes.
• The magnifying power of a simple magnifier increases as the focal length decreases (roughly m = 1 + D/f, with D ≈ 25 cm).
• The convex lens of focal length 5 cm gives much more magnification than the 50 cm one, so it is the better choice, option (c).

Answer: (c) A convex lens of focal length 5 cm, because a convex lens of short focal length gives the greatest magnification for reading small print.

A concave mirror gives an erect image only when the object lies between the pole and the principal focus (inside the focal length). The image is then virtual, erect and enlarged, formed behind the mirror. The focal length here is 15 cm.

• For an erect image the object must be between the pole P and the focus F, so the object distance must be less than the focal length: 0 < object distance < 15 cm.
• Nature of the image: the reflected rays only appear to meet behind the mirror, so the image is virtual and erect.
• Size: such an image is always larger (magnified) than the object.

For the ray diagram, draw one ray parallel to the axis (reflecting as if from F) and one ray towards F (reflecting parallel); their backward extensions meet behind the mirror to fix the enlarged virtual image.

Answer: Object distance: between 0 and 15 cm (between the pole and the focus). The image is virtual, erect and larger than the object.

The choice of mirror depends on what the situation needs: a concave mirror can make a strong parallel beam (source at its focus) or concentrate light to a hot spot, while a convex mirror gives an erect, diminished image over a wide field of view.

• (a) Headlights: concave mirror. The bulb is at the focus; light reflects as a strong parallel beam that travels far.
• (b) Side/rear-view mirror: convex mirror. It always gives an erect, diminished image and a wide field of view.
• (c) Solar furnace: concave mirror. A large concave mirror focuses parallel sunlight to a single point, producing intense heat.

Answer: (a) Concave mirror (strong parallel beam from a source at its focus). (b) Convex mirror (erect, diminished image with a wide field of view). (c) Concave mirror (concentrates sunlight to a hot focus).

Every point of an object sends out rays in all directions, and these rays strike the whole surface of a lens. A convex lens bends all of them to meet at the image point, so each part of the lens, by itself, can still form a complete image, just with fewer rays.

• Cover the upper (or lower) half with black paper; only the uncovered half now refracts.
• Rays from every point of the object still reach the uncovered half and bend to the correct image point, so a complete image of the whole object is still formed.
• Because only half the lens collects light, the image is less bright (fainter), but its shape, size and position are unchanged.

Verify on a bench with a candle, a convex lens and a screen, then slide a card over half the lens and watch the image stay whole but fade.

Answer: Yes, a complete image is formed even with half the lens covered. Only its brightness decreases (it becomes fainter), because fewer rays reach the image; its shape, size and position stay the same.

A converging lens is a convex lens. Use the lens formula 1/v - 1/u = 1/f and the lens magnification m = h'/h = v/u. By the sign convention u = -25 cm; the lens is convex, so f = +10 cm; object height h = +5 cm.

• Make 1/v the subject: 1/v = 1/f + 1/u.
• Substitute: 1/v = 1/10 + 1/(-25).
• Common denominator: 1/v = 5/50 - 2/50 = 3/50.
• Invert: v = 50/3 = +16.7 cm. Positive, so the image is on the far side: real.
• Magnification: m = v/u = 16.7/(-25) = -0.67.
• Image height: h' = m×h = (-0.67) × 5 = -3.3 cm. Negative means inverted; size 3.3 cm (smaller).

For the ray diagram, draw one ray parallel to the axis (refracting through F) and one ray through the optical centre (going straight); they meet to fix the small inverted image between F and 2F.

Answer: Image position v = +16.7 cm (on the far side of the lens), size ≈ 3.3 cm; the image is real, inverted and diminished.

A concave lens is diverging and always forms a virtual, erect, diminished image on the same side as the object. So both the focal length and the image distance are negative: f = -15 cm and v = -10 cm. Use the lens formula to find u.

• Make 1/u the subject: 1/v - 1/u = 1/f ⇒ 1/u = 1/v - 1/f.
• Substitute: 1/u = 1/(-10) - 1/(-15).
• Common denominator: 1/u = -3/30 + 2/30 = -1/30.
• Invert: u = -30 cm. The negative sign means the object is 30 cm in front of the lens.

For the ray diagram, draw one ray parallel to the axis (diverging as if from the focus on the object side) and one ray through the optical centre; their backward extensions meet to give the small upright virtual image.

Answer: The object is placed 30 cm in front of the lens (u = -30 cm).

A convex mirror is diverging; its focus is behind the mirror, so f = +15 cm. The object is in front, so u = -10 cm. Use the mirror formula 1/v + 1/u = 1/f.

• Make 1/v the subject: 1/v = 1/f - 1/u.
• Substitute: 1/v = 1/15 - 1/(-10).
• Common denominator: 1/v = 2/30 + 3/30 = 5/30 = 1/6.
• Invert: v = +6 cm. Positive, so the image is 6 cm behind the mirror: virtual and erect.
• Magnification: m = -v/u = -6/(-10) = +0.6. Positive and less than 1, so erect and diminished.

Answer: The image is 6 cm behind the mirror (v = +6 cm); it is virtual, erect and diminished (m = +0.6).

Magnification is the ratio of image height to object height, m = h'/h. Its sign tells whether the image is erect (+) or inverted (-), and its magnitude tells the relative size. A value of +1 carries two messages at once.

• Magnitude 1, so |h'| = |h|: the image is exactly the same size as the object.
• Sign positive, so the image is erect (upright), the same way up as the object.
• A plane-mirror image is always virtual, so m = +1 means virtual, erect and equal in size.

Answer: m = +1 means the image formed by the plane mirror is the same size as the object (magnitude 1) and erect (positive sign); it is also virtual.

For a convex mirror, f = R/2 and is positive. Use the mirror formula 1/v + 1/u = 1/f and magnification m = -v/u. Here R = +30 cm so f = +15 cm; u = -20 cm; h = +5.0 cm.

• Focal length: f = R/2 = +30/2 = +15 cm.
• Make 1/v the subject: 1/v = 1/f - 1/u.
• Substitute: 1/v = 1/15 - 1/(-20).
• Common denominator: 1/v = 4/60 + 3/60 = 7/60.
• Invert: v = 60/7 = +8.6 cm. Positive, so 8.6 cm behind the mirror (virtual, erect).
• Magnification: m = -v/u = -8.6/(-20) = +0.43.
• Image height: h' = m×h = 0.43 × 5.0 = +2.2 cm (erect, diminished).

Answer: Image position v = +8.6 cm (behind the mirror); nature: virtual, erect and diminished; size h' ≈ 2.2 cm.

A sharp image on a screen must be a real image, formed in front of a concave mirror. For a concave mirror f = -18 cm; u = -27 cm; h = +7.0 cm. Use the mirror formula and magnification.

• Make 1/v the subject: 1/v = 1/f - 1/u.
• Substitute: 1/v = 1/(-18) - 1/(-27).
• Common denominator (LCM 54): 1/v = -3/54 + 2/54 = -1/54.
• Invert: v = -54 cm. The negative sign means the image is 54 cm in front of the mirror, so the screen goes there. The image is real.
• Magnification: m = -v/u = -(-54)/(-27) = -2.
• Image height: h' = m×h = (-2) × 7.0 = -14 cm. Inverted; size 14 cm (twice the object, so enlarged).

Answer: Place the screen 54 cm in front of the mirror (v = -54 cm). The image is real, inverted and enlarged, of size 14 cm.

The power of a lens and its focal length are reciprocals, P = 1/f (f in metres), so f = 1/P. The sign of the power identifies the lens: positive means a convex (converging) lens, negative means a concave (diverging) lens.

• Write the relation: f = 1/P.
• Substitute P = -2.0 D: f = 1/(-2.0).
• Do the arithmetic: f = -0.5 m = -50 cm.
• A negative power (and negative focal length) means a concave (diverging) lens.

Answer: The focal length is -0.5 m (-50 cm), and the lens is a concave (diverging) lens (shown by the negative power).

Focal length is the reciprocal of power, f = 1/P (f in metres). A positive power means a convex lens, which is a converging lens; a negative power would mean a diverging (concave) lens.

• Write the relation: f = 1/P.
• Substitute P = +1.5 D: f = 1/(+1.5).
• Do the arithmetic: f = +0.667 m = +66.7 cm.
• A positive power (and positive focal length) means a convex, converging lens.

Answer: The focal length is +0.667 m (≈ +66.7 cm); the lens is converging (convex), shown by the positive power.